MONOMIAL HYPEROVALS IN DESARGUESIAN PLANES
by
Timothy L. Vis
B.A., Dordt College, 2005
M.S., University of Colorado Denver, 2008
A thesis submitted to the
University of Colorado Denver
in partial fulfillment
of the requirements for the degree of
Doctor of Philosophy
Applied Mathematics
2010
This thesis for the Doctor of Philosophy
degree by
Timothy L. Vis
has been approved
by
Timothy J. Penttila
Vis, Timothy L. (Ph.D., Applied Mathematics)
Monomial Hyperovals in Desarguesian Planes
Thesis directed by Professor William E. Cherowitzo
ABSTRACT
In a finite projective plane of order n, an oval is a set of n + 1 points such
that no three of the points are collinear. Each point of an oval lies on a unique
line tangent to the oval. When n is even the tangent lines to an oval meet in a
common point with which the oval can be extended to form a hyperoval. In the
plane PG(2,q) constructed from GF(q)3, irreducible conics provide examples
of ovals. Segre [32, 33] showed that when q is odd, these are the only examples.
The situation is quite different when q is even: numerous families and exam-
ples exist and classification seems intractable. The work of Glynn [11], however,
provides a powerful tool for classifying monomial hyperovals (those described
by monomials), suggesting that this may be a tractable problem.
Every known monomial hyperoval is equivalent to one described with an
exponent whose binary expansion has three or fewer nonzero bits. Work of
Segre [34] and of Cherowitzo and Storme [5] has fully classified the monomial
hyperovals described by two or fewer bits. We extend this work by classifying
the monomial hyperovals described by three bits.
Following certain reduction arguments, we use Glynns criterion [11] to an-
alyze possible exponents. The possibilities are restricted to eleven broad cases,
which are eliminated systematically through more narrowly focused application
of Glynns criterion.
With this classification complete, we discuss several avenues of research with
the goal of showing that any monomial hyperoval can be described with three or
fewer bits. We discuss alternate forms of monomial hyperovals, giving all forms
for the known monomial hyperovals and some of the alternate forms for arbitrary
one, two, and three bit exponents. We also discuss two means of transforming
the points of a projective plane that may be useful in simplifying monomial
hyperovals by viewing them in relation to other well-studied structures.
This abstract accurately represents the content of the candidates thesis. I
recommend its publication.
ACKNOWLEDGMENT
I would like to acknowledge Warren Batemans generous sponsorship of the
Lynn Bateman Memorial Doctoral Fellowship which allowed me four semesters
of release from teaching duties to concentrate on the work found in this thesis.
CONTENTS
Figures ................................................................ ix
Tables.................................................................. x
Chapter
1. MDS Codes and Arcs.................................................. 1
1.1 Error-Correcting Codes............................................ 1
1.2 The Singleton Bound and MDS Codes................................. 4
1.3 n-arcs in Projective Spaces....................................... 6
2. Background........................................................... 7
2.1 Projective Planes................................................. 7
2.2 Desargues Theorem and PG (2,K)................................... 9
2.3 Structure of Projective Planes................................... 11
2.4 Arcs, Ovals, and Hyperovals in PG (2,q).......................... 14
2.5 Monomial Hyperovals.............................................. 22
2.6 Proof of Glynns Criterion....................................... 24
3. Three-Bit Monomial Hyperovals....................................... 28
3.1 Reduction........................................................ 28
3.2 Common Divisor Conditions......................................... 30
3.3 Monomials With Exponent of the Form a + a1 + a?................... 40
3.4 Reducible Monomial Hyperovals..................................... 52
4. Proof of Theorem 3.29............................................... 55
vi
4.1 Case 1............................................................ 55
4.2 Case 2............................................................ 65
4.3 Case 3............................................................ 88
4.4 Case 4............................................................ 97
4.5 Case 5........................................................... 104
4.6 Case 6........................................................... 105
4.7 Case 7........................................................... 110
4.8 Case 8........................................................... 110
4.9 Case 9........................................................... 115
4.10 Case 10.......................................................... 117
4.11 Case 11 ......................................................... 150
4.12 Conclusion....................................................... 151
5. Projectively Equivalent Monomial Sets............................. 152
5.1 Alternative Forms of Monomial Hyperovals.......................... 152
5.1.1 Translation Hyperovals Q> (a)................................. 153
5.1.2 Segre Hyperovals ^ (6).......................................... 154
5.1.3 Glynn Hyperovals Q) (a + 7)..................................... 156
5.1.4 Glynn Hyperovals $) (3cr + 4)................................... 157
5.2 Some Equivalent Bit Patterns...................................... 158
5.2.1 One Bit Patterns................................................ 160
5.2.2 Two Bit Patterns................................................ 161
6. Relating Projectively Inequivalent Sets of Points................. 164
6.1 Approach from Cremona Transformations............................. 164
6.2 Generalized Oval Derivation....................................... 166
vii
6.2.1 Derivation...................................................... 167
6.2.2 Collineations Under Derivation.................................. 171
6.2.3 Deriving a Desarguesian Plane When P = Q..................... 175
6.2.4 Deriving a Desarguesian Plane When P ^ Q..................... 181
7. Future Research.................................................... 185
References............................................................ 187
viii
FIGURES
Figure
2.1 The Fano Plane...................................................... 8
2.2 A Coordinatized Fano Plane......................................... 10
2.3 Triangles in Perspective from the Point P.......................... 11
2.4 Triangles in Perspective from the Line l .......................... 12
2.5 The Desargues Configuration........................................ 13
2.6 An Oval and its Nucleus in the Fano Plane ......................... 16
IX
TABLES
Table
4.1 -Case 1 Classification Outline ............................................ 66
4.2 Case 2 Classification Outline ............................................ 88
4.3 Case 3 Classification Outline ............................................ 98
4.4 Case 4 Classification Outline ........................................... 105
4.5 Case 6 Classification Outline ........................................... 109
4.6 Case 8 Classification Outline ........................................... 115
4.7 Case 9 Classification Outline ........................................... 117
4.8 Case 10 Classification Outline.......................................... 148
x
1. MDS Codes and Arcs
1.1 Error-Correcting Codes
A crucial element of information transmission is the ability to send a message
and ensure, with high probability, that the same message will be received. When
noise or other factors cause errors in the transmitted message, the capability to
detect and correct those errors is critical to effective communication. For this
purpose, error-correcting codes are used. For a more thorough introduction to
the theory of error-correcting codes, see [29]. Here, we shall only consider codes
in which all codewords have the same length.
Example 1. One means of detecting and correcting errors would be to repeat
each letter three times. This is the three-repeat code. When a received word (of
three letters) does not consist of the same letter repeated three times, an error
is detected; however, if some letter appears twice, the word can be corrected to
that letter repeated three times.
Formally, we define a code in the following way.
Definition 1.1 Denote the space of all vectors of length n over an alphabet
of q elements as V (n, q). A code C over an alphabet of q elements and
having length n is simply a subset of M elements ofV (n,q). An element
of V (n, q) is a word, and an element of C is a codeword.
In the case of the three-repeat code of Example 1, codewords are of length
n = 3, and the code C consists of all words that contain the same letter repeated
three times.
1
Unless an error changes a transmitted codeword to another codeword, the
received word will not be a codeword, and the error will be detected. Further-
more, if there is a clear indication of what error was made, the received word
may be corrected back to the codeword that was originally transmitted. In order
to determine the correct codeword for a given received word, we define a metric
on V (n, q) as follows.
Definition 1.2 Given two words u and v, the Hamming distance d (u, v) is the
number of positions in which the two words differ. The minimum distance
d of a code C is the smallest distance between distinct codewords. A code
with length n, M codewords, and minimum distance d is an (n, M, d) code.
Again, referencing the three-repeat code of Example 1, the distance between
any two distinct words (and therefore the minimum distance) is three. Thus, in
an alphabet with q letters, the three-repeat code is a (3, q, 3) code.
The Hamming distance defines a metric on V (n,q). This metric may be
used to determine how many errors can be effectively detected or corrected with
the following theorem.
Theorem 1.3 In a code with minimum distance d, any d 1 or fewer errors
can be detected, and any t or fewer errors can be corrected.
Proof: At least d errors are required for a codeword other than that transmit-
ted to be received, so any d 1 or fewer errors can be detected. If t or fewer
errors occur, the distance to any codeword other than that transmitted is at
least t -f 1, and the transmitted codeword is the unique codeword closest to the
received word. Thus, any t or fewer errors can be corrected.
2
Applying Theorem 1.3 to the three-repeat code indicates that any two er-
rors may be detected and any one error corrected. Indeed, if the word aaa is
transmitted, two or fewer errors must leave at least one a and cannot yield an-
other codeword, while three errors might yield the codeword bbb, so that only
two errors may be detected. If only one error occurs, the received word will
look like baa, aba, or aab, and can unambiguously be corrected to aaa. If two
errors occur, the received word could look like abb or abc. In the first case, the
word would be improperly corrected to bbb; in the second the word cannot be
definitively corrected to any of aaa, bbb, or ccc. Thus only one error may be
properly corrected.
In general, an error-correcting code need not have any algebraic structure.
However, many of the most useful codes can be thought of algebraically and
the underlying algebraic structure can often be used to aid in efficient error-
correction. In particular, when q (the size of the alphabet) is a prime power we
may treat V (n, q) as the vector space GF (q)n and we may use some subspace
of GF (q)n as the code C. We shall restrict our attention to these linear codes,
which we now formally define.
Definition 1.4 A linear [n,k,d] code is a k-dimensional subspace of GF (q)n
with minimum distance d.
The three-repeat code of Example 1 over the alphabet GF (q) provides a
simple example of a linear code spanned by the vector (1,1,1). It is thus a
[3,1,3] code.
1.2 The Singleton Bound and MDS Codes
3
There is a natural tension between the parameters n, k, and d in terms of
constructing efficient codes that correct many errors. It is desirable that n k be
small, to minimize the amount of extra information transmitted. Put another
way, for a given n, it is preferable to maximize the number of codewords so as to
maximize the efficiency of transmission. At the same time, it is desirable that d
be large, to maximize the error-correcting capabilities of the code.
These two goals are naturally at odds: increasing the number of codewords
and increasing their mutual distance are incompatible goals. It is natural to ask
what the best possible situation could be. The Singleton Bound answers that
question, and the optimal codes, in terms of minimum distance, are those codes
that attain the Singleton Bound. The Singleton Bound can be used to bound
any of the parameters n, k, and d with the other two. For our purposes, we
express it as a bound on d.
Theorem 1.5 (Singleton Bound) d < n k + 1.
Proof: The Singleton Bound may also be written k < n d + 1. Consider
the first n d + 1 entries in the codewords of an [n, k, d]-code. Since only d 1
other entries exist, any two codewords must differ in at least one of these entries.
Thus, the dimension of the code, k, can be no more than n d + 1.
Codes which meet the Singleton Bound are those which attain the maximum
possible minimum distance for a given choice of n and k. We give these codes a
special name.
Definition 1.6 A code C for which d = nk+1 is called a Maximum Distance
Separable or MDS code.
4
The three-repeat code of Example 1 meets the Singleton Bound and is an
MDS code when the codewords are considered to have length three. However,
in a broader perspective, the three-repeat code becomes highly inefficient. If,
instead of thinking of encoding individual letters, we think of encoding words
of length n by repeating each letter three times, we obtain a [3n, n, 3] code, and
the minimum distance 3 is much less than the value 2n +1 required for the code
to be MDS. In general then, it is far more efficient to use other codes which
more closely approach the Singleton Bound, such as MDS codes if they exist.
MDS codes have a nice geometric interpretation [15, 36] and serve as a
significant motivator for the study of certain geometric structures. In order to
discuss this interpretation, however, we require a few more basic concepts related
to linear codes.
Definition 1.7 Given a linear code C, let G be a matrix with linearly inde-
pendent rows such that C = row (G) and let H be a matrix with linearly
independent rows such that C null (H). Then G is a generator matrix
for C and H is a parity-check matrix for C.
It is easy to see that for an [n, k, d]-code, a generator matrix is k x n and a
parity-check matrix is n k x n. In addition, a parity-check matrix for a code
can be used to determine its minimum distance as follows.
Theorem 1.8 A linear code with parity-check matrix H has minimum distance
d if every set of d 1 columns of H is linearly independent and some set of d
columns of H is linearly dependent.
5
Thus, any MDS code has aiJ-lxn parity-check matrix with every set of
d 1 columns linearly independent and any such matrix is a parity-check matrix
for an MDS code.
1.3 n-arcs in Projective Spaces
Consider then, the columns of a parity-check matrix for an MDS code over
GF (q). These columns form a set of n vectors in GF (q)dl in which every subset
of d 1 or fewer vectors is linearly independent.
These vectors may also be considered as coordinates of points in the pro-
jective space PG (d 2,q). Any set of m points with linearly independent co-
ordinates spans an m 1-dimensional projective space. Thus, these vectors
correspond to a set of n points with the property that any set of m < d 1
points spans anm-1 dimensional projective space, no m points lie in any m 2
dimensional projective space.
Definition 1.9 A set ofn points of PG (r, q) is an n- arc if for every m < r+1,
no set of m points lies in any r l-dimensional subspace of PG (r, q).
It should be clear from this definition that the columns of a parity-check
matrix for an MDS code coordinatize the points of an n-arc; the converse is also
true. As such, the study of arcs in projective spaces (of any dimension) is of
interest in determining optimal codes.
6
2. Background
2.1 Projective Planes
In order to discuss monomial hyperovals in PG (2,2h), we must first discuss
what PG (2, K) is. We begin by defining a projective plane and discussing
certain important properties of projective planes.
Definition 2.1 A projective plane is a point-line incidence geometry satisfying
the following three axioms:
1. Any two distinct points are incident with a unique common line.
2. Any two distinct lines are incident with a unique common point.
3. There exists a set of four points such that no three are incident with
a common line.
The last axiom is a non-degeneracy axiom that excludes certain undesirable
and uninteresting cases.
Example 2. The smallest projective plane, pictured in Figure 2.1 has seven
points and seven lines. Note that the circle in the figure represents a line in this
plane. Each point is incident with three lines, and each line is incident with
three points.
Before we proceed further, we introduce several concepts.
Definition 2.2 A set of points in a projective plane is collinear if there exists
a line incident with each of the points. A set of lines in a projective plane
is concurrent if there exists a point incident with each of the lines.
7
Figure 2.1: The Fano Plane
Definition 2.3 A skewfield consists of a set S and two operations + and -,
such that (S, +) is an Abelian group with identity 0, (S \ {0} ) is a group,
and a -(b + c) = a- b + a- c for all a, b, and c (not necessarily distinct) in
S. In other words, a skewfield satisfies all of the axioms of a field except
possibly commutativity of multiplication.
Higher dimensional projective spaces can all be constructed from vector
spaces over skewfields; the same cannot be said for projective planes. In fact,
the classification of all projective planes remains perhaps one of the most out-
standing and significant open problems in combinatorics. Our interest, however,
shall be solely in those projective planes constructed from vector spaces over
skewfields: the planes PG (2, K).
2.2 Desargues Theorem and PG (2, K)
8
Definition 2.4 The geometry PG (2, K) is the point-line incidence structure
whose points are the one-dimensional vector subspaces of K3, lines are
the two-dimensional vector subspaces of K3, and incidence is the natural
subspace relation. When K is the finite field GF (q), we simply write
PG (2, q).
The geometry PG (2, K) is, as alluded to earlier, a projective plane. Proving
this is a simple exercise in linear algebra.
Theorem 2.5 The geometry PG (2, K) is a projective plane.
PG (2, K) has the desirable property that it can be easily coordinatized
with both points and lines receiving coordinates. We will represent points of
PG (2, K) as non-zero vectors in the corresponding space with parentheses and
lines of PG (2, K) as non-zero vectors in the orthogonal space with brackets. In
each case, two sets of coordinates that are scalar multiples of one another refer
to the same point or line. In Figure 2.2, we present such a coordinatization of
the Fano plane using GF (2).
In addition to the algebraic structure that defines PG (2, K), there is also a
geometric property that distinguishes these planes: they are precisely the planes
that satisfy Desargues Theorem. In order to describe Desargues Theorem, we
require the following definitions.
Definition 2.6 A triangle is a set of three non-collinear points. Triangles
QRS and Q'R!S' are said to be perspective from a point or central if
the lines QQ', RR', and SS1 are concurrent (see Figure 2.3). Triangles
9
Figure 2.2: A Coordinatized Fano Plane
are said to be perspective from a line or axial if the points QR D Q'R',
QS fl Q'S', and RS fl R'S' are collinear (see Figure 2.4).
Definition 2.7 A projective plane satisfies Desargues Theorem and is called
Desarguesian if and only if every pair of triangles that is perspective from
a point is perspective from a line. Figure 2.5 shows triangles QRS and
Q'R'S' that are perspective from the point P and from the line l in the
Desargues configuration.
The Fano plane is somewhat exceptional in that it does not contain any
pairs of disjoint triangles. In fact, the Desargues configuration (pictured in
Figure 2.5) contains ten points, and the Fano plane only seven. If, however, we
10
Figure 2.3: Triangles in Perspective from the Point P
do not require that the triangles are disjoint, any triangles in the Fano plane
that are perspective from a point are perspective from a line. If we do require
that the triangles are disjoint, the statement that any pair of disjoint triangles
that are perspective from a point are also perspective from a line is vacuously
true in the Fano plane. As such, we consider the Fano plane to be Desarguesian.
2.3 Structure of Projective Planes
In order to further our study of projective planes, we now turn to an exami-
nation of the structure of a projective plane. The following result, which is true
for any projective plane, establishes a great deal of structure within a projective
plane.
11
Figure 2.4: Triangles in Perspective from the Line l
Theorem 2.8 In a projective plane, given points P and Q and lines l and m,
there are bijections between each of the following: lines through P, lines through
Q, points on l, and points on m. In particular, in a finite projective plane, there
is some number n such that every line contains n + 1 points, every point lies on
n + 1 lines, and such that the plane contains n2 + n + 1 points and the same
number of lines. m
Definition 2.9 The number n described in Theorem 2.8 is called the order of
a plane.
Theorem 2.8 indicates that from the simple angle of counting the number
of lines incident with a point (or points incident with a line), all points (or all
lines) are equivalent. It is natural to consider when points or lines are equivalent
from a more meaningful perspective. The next definition describes the analogue
to an isomorphism in incidence geometry: a map that preserves the incidence
relations.
12
Figure 2.5: The Desargues Configuration
Definition 2.10 A bijective map a on the points and lines of a projective plane
is called a collineation of the plane provided that whenever a point P is
incident with a line l, the point Pa is incident with the line la.
It is not difficult to see that the set of all collineations of a projective plane
forms a group, called the full collineation group of the projective plane. When
the projective plane is PG (2, K), the structure of this group is known. Two
families of collineations form the basis for this group.
Definition 2.11 A homography is a collineation of PG (2, K) induced by mul-
tiplication of each point by an invertible 3x3 matrix.
13
Definition 2.12 An automorphic collineation is a collineation induced by ap-
plying the same automorphism of K to each coordinate of each point.
Theorem 2.13 (Fundamental Theorem of Field Planes) If F is a field,
every collineation of PG (2, F) is the product of a homography and an automor-
phic collineation.
2.4 Arcs, Ovals, and Hyperovals in PG(2,q)
In 1947, Bose [3] studied the application of finite geometries to the theory of
confounding in factorial designs. One of the particular questions he addressed
was the construction of symmetric factorial designs in which interactions of a
certain degree (t) or lower remained unconfounded. He was able to construct
such designs using sets of points in the projective space PG (r, q) such that every
t of them span a subspace of dimension t 1.
Notice that when t = r, such a set of points is an n-arc. Because our interest
from this point forward will lie strictly within projective planes, we redefine an
arc within this setting.
Definition 2.14 A k-arc (or simply an axe) in a projective plane is a set of k
points, no three collinear.
Theorem 2.15 (Bose, [3]) The maximum size of an arc in PG (2, q) is q + 2
when q is even and q + 1 when q is odd.
Given an arc JF in an arbitrary projective plane 7r, the lines of 7r can be
separated into three classes by the size of their intersection with JF. Lines not
14
intersecting are external lines; lines intersecting in a single point are
tangent lines; lines intersecting Jtf in two points are secant lines. Qvist was
able to extend Boses result to arbitrary projective planes in the following way.
Theorem 2.16 ([30]) In a plane of order q, if q is even the tangent lines to a
q + 1 arc all meet in a common point (called the nucleus); if q is odd every point
on a tangent line to a q + 1 arc lies on some other tangent line.
Corollary 2.17 In a plane of order q, every q +1 arc can be extended to a q + 2
arc when q is even, and no q + 2 arcs can exist when q is odd.
Our interest is in these arcs of maximum size, for which we make the fol-
lowing definition.
Definition 2.18 A q + 1 -arc in a projective plane of order q is an oval; a
q + 2-arc is a hyperoval.
Example 3. Figure 2.6 illustrates an oval in the Fano plane. Lines k, l, and
m are, respectively, external, tangent, and secant to the oval, while the point N
is its nucleus.
In the more general case of PG(2,q), an example of an oval is given by
the points whose coordinates satisfy an irreducible quadratic equation. When
q is even, extending this set of points yields an example of a hyperoval. These
examples are named in the following definition.
Definition 2.19 The set of points in PG (2, q) whose coordinates satisfy an
irreducible quadratic equation is a conic or regular oval. If q is even, the
extension of this set of points is a hyperconic or regular hyperoval.
15
k
l \m
Figure 2.6: An Oval and its Nucleus in the Fano Plane
In 1955, Segre proved his celebrated theorem, cementing the connection
between the combinatorially described oval (set of q+l points, no three collinear)
and the algebraically described conic (set of points satisfying an irreducible
quadratic equation).
Theorem 2.20 (Segre, [32, 33]) Every oval in PG(2,q), q odd, is a conic.u
When q is even, the situation is quite different, and a complete classification
of ovals in PG(2,q), q even seems at present to be intractable. A number of
families have been constructed, which we discuss after a few more preliminary
definitions and results.
Definition 2.21 The points (0,0,1), (0,1,0), (1,0,0), and (1,1,1) are to-
16
gether called the fundamental quadrangle.
A consequence of the Fundamental Theorem (Theorem 2.13) is that given
any two ordered quadrangles of PG(2,q), there exists a collineation mapping
one to the other in the same order (essentially a change of basis). This gives us
the following equivalence.
Theorem 2.22 Every hyperoval in PG (2, q) is projectively equivalent to one
containing the fundamental quadrangle.
We may assume, therefore, that a hyperoval contains the fundamental quad-
rangle and consider the implications for the remaining points. Notice that any
point of the form (0, x, y) lies on the line joining (0,1,0) and (0, 0,1), so that the
remaining points all have non-zero first coordinate. Without loss of generality,
the remaining points have coordinates (1, x, y). Now if any two such points have
the same second (or third) coordinate, the point (0, 0,1) (or (0,1,0)) lies on the
line joining those two points. Thus, any two such points have distinct second
and third coordinates. Since there are q such points, we may assume these points
are of the form (1, x, / (x)), as x ranges over the elements of GF (q), and where
/ is a permutation polynomial. We define, more generally, the following set of
points.
Definition 2.23 The set of points Q) (/) is the following set:
(f) = {(l,x,f(x))\xeGF (<,)} U {(0,1,0), (0,0,1)} .
When f (x) xk, we will simply write (k) for Td (/).
17
Following Cherowitzo [6], we call any function / with / (0) = 0 and / (1) = 1
for which 2 (/) is a hyperoval an o-polynomial.
Considering the lines through the point (1, Â£,/(Â£)) and any other point of
the hyperoval, we also notice that the values are distinct if and only
if the lines are distinct. Thus, we obtain the following criterion, due to Segre
[34, 35] for a polynomial to be an o-polynomial.
Theorem 2.24 ([34, 35]) The permutation f is an o-polynomial if and only if
the following three conditions hold:
1. /(0) = 0,
2. /(1) = 1,
3. Each ft (x) is a permutation polynomial, where
X + t
We should note that the set 2 (/) is a hyperoval whenever the third condi-
tion holdsthe requirement that / (0) = 0 and / (1) = 1 simply forces (/) to
contain the fundamental quadrangle. In this notation, conics can be described in
several ways, most notably as (2). With this notation, we are able to discuss
the other known examples of hyperovals.
Two years after proving his famed theorem, Segre [34] constructed the trans-
lation hyperovals using field automorphisms of maximum order. These hyper-
ovals have the form 2 (21), (i, h) = 1 and exist distinctly from regular hyperovals
18
in PG (2, 2h) for h = 5 and h > 7. In the same paper, Segre established that
all hyperovals in PG (2, 2h), h < 3 were regular (projectively equivalent to con-
ics) and asked whether any irregular (non-conic) hyperovals existed in the two
remaining cases.
For PG (2,24), the answer came quickly when Lunelli and See [19] con-
structed an irregular hyperoval with o-polynomial 7712x2 + r710x4-f773x8 + r/12a:10-|-
rfx12 + r]4xu, 77 6 GF (24) with rj4 77 + 1. For PG (2, 26) the answer was much
longer in coming and several other families and examples appeared in the inter-
vening time.
Following these hyperovals, many further examples and families were found.
In 1962, Segre announced the construction of the Segre hyperoval 2 (6) in
PG (2, 2h), h odd [35]. The proof, however, did not appear until 1971 [31].
A few years later, Eich, Payne, and Hirschfeld [10, 14] showed that 2 (20) was
a hyperoval in PG (2, 27) distinct from the translation and Segre hyperovals.
With the aid of a computer, Hall [13] showed in 1975 that the only hyperovals
in PG (2, 24) were the regular hyperoval and the Lunelli-Sce hyperoval; in 1991,
OKeefe and Penttila [21] provided a computer-free proof of the result.
Glynn [11] introduced two further infinite families in 1983: 2 (3cr + 4) and
2) (cr + 7) in PG (2, 2h), h odd, where cr2 = 2 = q4 mod 2h 1. Glynns hyper-
ovals included the earlier example of [10, 14]. At the same time, he introduced
a computationally advantageous criterion for determining when 2 (k) is a hy-
peroval. As this criterion is critical to our work, we prove it at the end of this
chapter.
Definition 2.25 Define the partial order < as follows. Write x = Y^ox^1
19
and y = YlrLoVi^ so that each %i G {0,1} and each {0,1}. Then
x < y if and only if Xj < yi for all i. In other words, x ^ y if and only if
in each position in the binary expansion of x containing a one, the binary
expansion of y also contains a one.
Example 4. Since 11 = 2 + 21 + 23, 12 = 22 + 23, and 13 = 2 + 22 + 23,
11 2< 13, but 12 X 13.
Theorem 2.26 (Glynns Criterion [11]) f (t) = tk is an o-polynomial over
PG (2, q) if and only if d H kd, for all 1 < d < q 2, where kd is reduced modulo
q 1 under the convention that zero is reduced to zero and any other multiple
of q 1 is reduced to q 1.
In 1985, Payne [24] discovered the first infinite family of non-monomial
hyperovals with o-polynomial xs +x5 -frrl in PG (2, 2^), h odd, while studying
generalized quadrangles. Three years later in 1988, Cherowitzo [6] discovered the
initial examples of the Cherowitzo hyperovals with h odd, having o-polynomial
xa + xa+2 + x3tT+4, where a2 = 2 mod 2h 1. In 1998 [7], he showed this to be
an infinite family.
A few years later, Glynn [12] extended his criterion to arbitrary polynomials
as follows:
Theorem 2.27 ([11]) If f (0) = 0 and /(1) = 1, @ (/) is a hyperoval if and
only if for all pairs of integers (b, c) with l**
the coefficient on xc in [f (x)]b is zero. **
20
In 1992, OKeefe and Penttila, examining possible stabilizers of hyperovals
in PG (2,25), produced a new irregular (non-conic) hyperoval with o-polynomial
x4 + x16 + x28 + rjn
(X6 + X10 + Xl4 + X18 + X22 + X26)
W (x8 + x20)+v6 (x12 + x24).
Two years later, Penttila and Pinneri [25] produced the first irregular hyper-
oval in PG (2, 26). Subsequently, Penttila and Royle [26, 27] produced further
examples of irregular hyperovals in PG (2, 2h) with 6 < h < 8. They further
proved that the only hyperovals in PG (2, 25) are the known ones, and that any
unknown hyperoval in PG (2, 26) has a trivial stabilizer.
The 1990s saw the development of techniques using flocks of cones to con-
struct o-polynomials, and Cherowitzo, Penttila, Pinneri, and Royle used flocks
of the quadratic cone to construct the two families of Subiaco hyperovals in
PG (2,2h). These have o-polynomials given as follows: Let g be an element of
GF (2h) such that t/-1 has trace one and 1 4- 77 + rj2 ^ 0. Let
g(x) = 12 +
rfx4 + rf (1 + 77 + 772) x3 + rf (1 + 77 + rf) x2 + rfx
(x2 + gx + l)2
and let
, _ 772 1 774x4 + ?73 (1 + 77 + T}2)2 x3 + rf (1 + t?2) x
77 (1 + 77 + 772) 77 (1 + 77 + 77s) (x2 + 77X + l)2
Then the o-polynomials are g (x) + tf (x) + t^x^, t E GF (2fc) and / (x). This
infinite family includes the Lunelli-Sce hyperoval.
Further new examples were developed in 1997 by Payne, Penttila, and Royle
[23] in the planes PG (2, 2l) for i E {10,12,14,16}.
21
In [7], Cherowitzo developed the theory to use flocks of cones over trans-
lation ovals to construct hyperovals, allowing him to establish the Cherowitzo
hyperovals as an infinite family. Five years later, Cherowitzo, OKeefe, and
Penttila [8] constructed the Adelaide hyperovals in PG (2, 2fe), h even with this
technique. These hyperovals have o-polynomial
T(ym)(t+1) , r((>t + >2) ) ,
T(,) T(ri)(t+ T(rnt< I l)" '
where rj E GF (2h) with r] ^ 1 and rj1'2+l = 1, T (x) = x + x, and m =
Together, the Adelaide and Subiaco hyperovals include each of the hyper-
ovals of [23], [25], [26], and [27], thereby placing every known hyperoval ex-
cept the OKeefe-Penttila in an infinite family. The known hyperovals then
are the translation hyperovals (including the regular hyperovals), Segre hyper-
ovals, Glynn hyperovals (two families), Payne hyperovals, Cherowitzo hyper-
ovals, Subiaco hyperovals, Adelaide hyperovals, and the OKeefe-Penttila hy-
peroval.
2.5 Monomial Hyperovals
Of the known hyperovals, the first four families discovered (translation hy-
perovals, Segre hyperovals, and both families of Glynn hyperovals) have mono-
mial o-polynomials. These hyperovals have, in general, a much greater degree
of symmetry than their non-monomial counterparts and lead to further combi-
natorial structures, making them an interesting subject of study.
One of the fundamental properties of a hyperoval is its stabilizer group in the
full collineation group of the ambient projective plane. In 1978, Korchmaros [17]
22
classified those hyperovals having a stabilizer acting transitively on the points
of the hyperoval.
Theorem 2.28 ([17]) The only hyperovals having a transitive stabilizer are the
regular hyperovals in PG (2,2) and PG (2,22) and the Lunelli-Sce hyperoval in
PG (2,24).
In 1994, OKeefe and Penttila [22] extended these results, considering hy-
perovals with large point orbits under the full collineation group of the plane.
Theorem 2.29 ([22]) The only hyperovals containing a transitive q + l-arc are
the regular hyperovals.
Theorem 2.30 ([22]) The only hyperovals containing a transitive q-arc are the
translation hyperovals.
Theorem 2.31 ([22]) The only hyperovals containing a transitive q 1-arc are
the monomial hyperovals unless q 212m+4.
As the transitive hyperovals, ovals, and q-arcs are thus completely charac-
terized (since regular and translation hyperovals are completely characterized),
a logical next step appears to be a complete characterization or classification
of monomial hyperovals. Along with a characterization of the q = 212m+4 case,
this would provide a complete classification of arcs with large point orbits under
their stabilizer groups.
Monomial hyperovals are also of interest because of their connection to cyclic
difference sets. Maschietti [20] has shown that whenever @ (k) is a hyperoval in
PG (2, 2^), the set {xk + x\x G GF (2ft) } is a difference set in Z2h_i- Because
23
of the interest in difference sets, a classification of monomial hyperovals is again
desirable.
A natural method of classification suggests itself from Glynns Criterion
(Theorem 2.26). This method considers the number of ones in the binary ex-
pansion of k, and some progress has been made in this direction. Segres work
in [34] completely characterizes all hyperovals of the form (2l). More re-
cent work of Cherowitzo and Storme [5] characterizes the hyperovals of the form
@ (2* + 2J). Since every known monomial hyperoval can be represented in either
one, two, or three bits, a two stage classification program is suggested:
1. Classify hyperovals of the form 3> (2l + 2n + 212).
2. Show that every monomial hyperoval can be represented in at most three
bits.
With extensive use of Glynns Criterion, we are able to complete the first part
of this program; the second remains an ongoing area of research.
In addition, Glynns Criterion allows for computational ease in classifying
monomial hyperovals in small planes. In [12], Glynn classified the monomial
hyperovals in PG (2,2h), for h < 30 as the known examples. Private com-
munication indicates that unpublished work has extended this classification to
considerably higher values of h. This suggests that new monomial hyperovals
are unlikely to exist.
2.6 Proof of Glynns Criterion
Because of its fundamental importance to our work, we now proceed to
provide a proof of Glynns Criterion. We begin with necessary preliminary re-
24
suits, starting with a monomial adaptation of Segres condition on o-polynomials
(Theorem 2.24).
Lemma 2.32 ([31, 35]) @ (k) is a hyperoval if and only if the following three
conditions hold: (k,q l) = 1, (k l,q 1) = 1, and ^-!l! permutes
GF(q)*.
Proof: The last condition requires k ^ 0. Thus, 0fc = 0 and \k = 1. Now xk is
a permutation if and only if (k, q 1) = 1 so by Theorem 2.24, we have only to
show that ***** is a permutation for each t. When t = 0, ^ is a permutation if
and only if (A: l,q 1) = 1. When t 0, is a permutation if and only if
(x+t)k+tk
is a permutation if and only if t
fc(iiri;
is a permutation if and only
r (x+l)k + l
if
is a permutation, establishing the result.
We also require certain algebraic results regarding polynomials over finite
fields.
Lemma 2.33 ([9]) The polynomial of degree at most q 1 representing a
function f (x) on a field has coefficients given as follows a0 = /(0), =
- ZteGFfo)* / (*) f~r 1
Theorem 2.34 (Hermite-Dickson Criterion [18]) Let GF (q) be a field of
characteristic p. Then f G GF (q) is a permutation polynomial of GF (q) if and
only if the following two conditions hold:
1. [f (f)]r has degree at most q 2 modulo tg t for r with 1 < r < q 2,
and r ^ 0 mod p.
2. f has exactly one root in GF (q).
25
Lemma 2.35 The expansion of (1 + x)k modulo xq x over GF (q) is given by
x md X<1 ~ x> where 0 < c < q 1.
With these results, we are prepared to prove Glynns Criterion.
Theorem 2.36 (2.26[11]) f (t) = tk is an o-polynomial over PG (2,q) if and
only if d ^ kd, for all 1 < d < q 2, where kd is reduced modulo q 1 under the
convention that zero is reduced to zero and any other multiple of q 1 is reduced
to q 1.
Proof: Let k be given as desired. Then by Lemma 2.32, xk is an o-polynomial
if and only if fk (x) = ^x+1^ -1 is a permutation. But ^X+1J +1 = ^z+i)+i > w^ich
simplifies to ^2i=o (x + l)*- In particular, fk (0) is simply the sum of k ones
and is zero or one as k is even or odd. For x ^ 0, fk (x) = 0 if and only if
(x + l)fc +1 = 0 if and only if (x + l)fe = 1 if and only if x +1 = lfc_1 if and only
if x 0, a contradiction. So to be a permutation, k must be even (to obtain
a root) and [fk (x)]r modulo xg x must have zero coefficient on xq~l for all
l
By Lemma 2.33, this coefficient is
- E
teGF(q)* V
for all 1 < r < q 2.
Now let g (x) = ^(x + l)fc + 1 j There is a unique polynomial / (x) repre-
senting this of degree at most q 1, and by Lemma 2.33, / (x) has coefficient on
26
xr of X^teGF(g)* ^t+1f 'fl j Thus, fk (x) is a permutation if and only if the
coefficient of xr in + l)k + 1 j is zero for all 1 < r < q 2.
Equivalently, the expansion Yhd giyen by Lemma 2.35 has zero
coefficient on xr for all 1 < r < q 2. But the terms with an xr are exactly
J2d^r Ylr
is equivalent to the statement that \{d : d ^ r ^ kd}\ = 0 mod 2 for all r.
With this given, suppose that there exists some d' such that d! -< kd! and
such that d' is minimal. Then r = d' yields {d \ d < d' -< kd) {d1}, a contra-
diction, so that this occurs if and only if d ^ kd for all d.
Now d kd for all d implies (k, q 1) = (k 1, q 1) = 1 in the following
way. If (k, q 1) = l > 1, m = satisfies mk = q 1 mod q 1. Then
m < mk, a contradiction, so that (k,q 1) = 1. If (k 1, q 1) = n ^ 1,
p = so p (k 1) = 0 mod q 1, which implies that pk = p mod q 1
giving them the same binary expansions, again a contradiction. So d ^ kd for
all d if and only if the three conditions of Lemma 2.32 are all satisfied.
27
3. Three-Bit Monomial Hyperovals
3.1 Reduction
To classify monomial hyperovals in PG (2,2h) of the form Q) (2* + 211 + 212),
it is helpful to restrict our attention to a more limited class of exponents in an
attempt to classify these hyperovals.
Definition 3.1 Let k = ]T)"_02a*, with a, < h. Further, let h = be. If, for
every pair {i,j}, ai ^ aj mod b, we say that k is 6-reducible with respect
to h. The number k' = $^_02ai, where a' is the reduction modulo b of aj
to lie in [0, 61], is called the 6-reduction of k with respect to h. If k is
not b-reducible, we say k is 6-irreducible with respect to h.
More generally, we make the following definition.
Definition 3.2 If there exists a 6 such that b\h and k is b-reducible, we say that
k is reducible with respect to h. If, on the other hand, k is b-irreducible
for every divisor 6 of h, we say that k is irreducible with respect to h.
When we are working in a finite field of order 2h, we say reducible and
irreducible for reducible with respect to h and irreducible with respect to h re-
spectively.
The idea of reducibility allows a significant lessening of the scope of the
necessary classification, as the following results will indicate. Our first result
shows that if a 6-reduction of k does not determine a monomial hyperoval, k
does not determine a monomial hyperoval.
28
Theorem 3.3 Let h = be and let k' be a b-reduction of k with respect to h. If
(k1) is not a monomial hyperoval in PG (2,2b), then 3> (k) is not a monomial
hyperoval in PG (2, 2h).
Proof: Suppose (kr) is not a monomial hyperoval in PG(2,26). Then
there exists some d' such that 1 < d' < 2b 2 and d' < k'd' mod 2b 1. Let
d = YfiZo 2bxd'. Now 1 < d < 2h 1, so it remains only to show that d z< kd
mod 2h 1.
The binary expansion of the value of d we constructed consists precisely of
the binary expansion of d! repeated c times. As such, 2bid = d mod 2h 1
for any integer i. More generally, 2bl+ad = 2ad mod 2h 1 for any choice of i
and ait is only the congruence class of a mod b that determines the product
2ad. Since the congruence classes of the bits in k coincide with the congruence
classes of the bits in k', then kd = k'd mod 2h 1. Thus, kd = YliZo 2bik'dl
mod 2h 1. Since d is simply d' repeated c times and kd is simply k'd' repeated
c times, we must have d < kd, so that & (k) is not a monomial hyperoval in
PG (2,2h). m
This leads naturally to the question of when a potential hyperoval exponent
is reducible, and motivates the classification of irreducible exponents and of
those hyperovals having irreducible exponents. It follows immediately from the
work above that any ^-reduction of an exponent yielding a monomial hyperoval
must itself give a monomial hyperoval. This suggests the following method of
classification:
1. Classify all monomial hyperovals having irreducible exponents.
29
2. Use the classification of hyperovals having irreducible exponents to classify
monomial hyperovals that have reducible exponents.
In the case that k = 2l + 211 + 212 we may easily restrict the occurrence of
irreducible exponents.
Theorem 3.4 If k = 210 + 2n + 212 andp, q, and r are distinct primes dividing
h, then k is reducible.
Proof: Suppose h = paq^s, where p \ s, q\ s, and s > 1. Further suppose that
k = 2l + 2n + 212 and that k is irreducible. Since k is irreducible, any divisor
of h must divide at least one of the differences between distinct bits in k or
a reduction with respect to that divisor would necessarily exist. In particular,
paqP must divide one 0f z2 ix, U z0, and ?'0 i2. Without loss of generality,
then, paq0\io ii-
Again, since k is irreducible, pas must divide one of the three differences.
However, ifpQs|i0 i2, it must be that h\i0 z2, a contradiction. Thus, pas\ii i0
or pas|i2 i\. Without loss of generality, then, pas\i\ Iq. But then, since
pa\i0 i2 and pa|ii i0, pa\i2 i\.
Finally, since k is irreducible, q@s must divide one of the three differences.
However, if q^s\ia since pa\ia h, h\ia % So q^s cannot divide any of the
three differences, a contradiction, so that k must be reducible.
3.2 Common Divisor Conditions
If k = 2l + 211 + 212 it is helpful if we can normalize with respect to one of
the exponents. If, in fact, one of the exponents (without loss of generality io)
is relatively prime to /i, a convenient normalization is possible, allowing us to
30
adapt Glynns criterion to an oary expansion, where a = 2l, since the a-ary
expansion is simply a permutation of the indices on the binary expansion. Our
main result is that for an irreducible exponent k = 2l + 2n + 2*2, this is always
possible. Our first few results, however, do not require that k be irreducible.
The results that follow through the remainder of this chapter and the next
make extensive use of Glynns criterion to show that under certain conditions
on k and h, (k) is not a monomial hyperoval in PG (2,2h). Since Glynns
criterion will be used in the same manner throughout, it will be helpful to explain
the procedure for using Glynns criterion.
In each case, we shall choose some value of d and use this value to compute
the product kd. Since k = 2l + 211 + 2*2, this is accomplished by distributing
the product as 210d + 2Hd + 212d. In order to show that d ^ kd, thereby showing
that @ (k) is not a monomial hyperoval in PG (2, 2h), two conditions must be
satisfied. Every bit of d must appear in one of the three summed terms, and
none of the three summed terms can coincide, resulting in carries. This is, of
course, the simplest situation. In many cases some bits of d will not appear in
any of the summed terms, and some of the summed terms will coincide. When
some bits of d do not appear, we will always show that carries from terms which
coincide will produce those bits of d. When other terms coincide, we will also
show that they do not produce any bits of d. In essence, we examine all of the
terms which coincide and show that after performing the necessary carries, and
under the stated conditions, d < kd.
This procedure is used in two primary ways. In the initial results, a gen-
eral value of d is chosen and examined to determine which values of k can be
31
excluded with the value of d. This procedure excludes wide swaths of values of
k and provides a narrow focus for the remaining work. The remaining work, in
later results, typically starts with a specific value or family of values of k and
determines a value of d which excludes that value of k.
Our work with Glynns criterion requires that everything be reduced
mod 2h 1. Throughout, we represent values that are congruent mod 2h 1
as equal, allowing us to eliminate the visual clutter of constantly writing
mod 2h 1 and allowing us to reserve the use of congruence for instances when
some other modulus is employed.
Lemma 3.5 Suppose m > 1, and m divides each ofi0) i\, i2; and h. Then if
k = 2l + 2n +212, Q) (k) is not a hyperoval in PG (2, 2.
Proof: Let d = 2J'TO. Then kd = 3 2jm = + 2jm+1).
Since jm + 1 ^ am for any a, d -< kd and 1 < d < 2h 2. Thus, (k) is not a
hyperoval in PG (2, 2h).
We further restrict the form of k with the following lemmas.
Lemma 3.6 7/(z0, h) > 1, (*i, h) > 1, (z2, h) > 1 and there exists no m dividing
each ofio, i\, ?2 and h, then for some ij, there exists an n dividing ij and h, but
not dividing ii for any l ^ j.
Proof: Since there is no m dividing each of the ij, the numbers (i0, h), (i\, h),
and (i2,h) cannot all be equal. Choose a = max {(i0, h), (*i, h), (i2, h)}. If a
occurs only once as (ij, h), then n = a divides ij and h, but not b for any l ^ j.
On the other hand, if a occurs twice, then let (ij,h) 7^ a, so that either
(ij, h) \ a or (ij, h) is a common divisor of each and h greater than 1. Since
32
this is excluded by the hypothesis, (ij, h) \ it for / ^ j, and n = (ij) satisfies the
conditions of the lemma.
We now assume, without loss of generality that a number m exists dividing
io and h but neither i\ nor i2.
Lemma 3.7 If ii and i2 are not both congruent to 1 mod m and k = 2l +
2n + 212, (k) is not a hyperoval in PG (2, 2ft).
Proof: Let d = Then kd = (2^m + 2-?m+ri + 2-?m+r'2), where
r\ and r2 are the reductions of i\ and i2 mod m to lie in [0, m 1], If ri and
r2 are not both m 1, 2jm+ri + 2^m+r2 ^ 2^+1^m, so that d ^ kd and @ (k) is
not a hyperoval in PG (2, 2h).
In the case that (i0, h), (i\,h), and (i2, h) are all distinct, we can strengthen
this condition considerably.
Lemma 3.8 Suppose (i0,h), (ii,h), and (i2,h) are distinct and each greater
than one. Then = i2 = 1 mod (i0,fi), i0 = i2 = 1 mod (ii,fi), and
i0 = ii = 1 mod (i2, fi) or ^ (fc) is not a hyperoval in PG (2,2h).
Proof: Lemma 3.7 guarantees that, without loss of generality, i\ = i2 = 1
mod (i0,fi). Without loss of generality, (ii,fi) does not divide (i2,h) and thus
does not divide i2. Further, since i\ = 1 mod (i0, h), (i\,h) does not divide
(i0, h) and thus does not divide i0. So, applying Lemma 3.7 again, i0 = i2 = 1
mod (ii, h). But then since i2 = 1 mod (i0, h), (i2, h) does not divide io, and
since i2 = 1 mod (ii, h), (i2, h) does not divide i\. So Lemma 3.7 now implies
that io = h = 1 mod (i2,fi).
33
We are left with two possibilities: either all three of (i0, h), (ii, h), and (i2, h)
are distinct, or (i\,h) = (i2,h) = 1 mod (io,h). We are able to rule out the
latter case with a series of results that mimic similar results of Cherowitzo and
Storme in their classification of the monomial hyperovals of the form (2* + 2l1)
[5],
Lemma 3.9 Let k = 2l + 2n + 212, where 2| (ii,h) and 2| (i2,h) and where
2 f (io, h). Then (k) is not a hyperoval in PG (2, 2h).
Proof: Notice that io is necessarily odd, since i\ is even and congruent to
-1 mod (i0,h). Now let d = eIo2^'1- Then kd = 2EL^2i~l + Ezio22'-
Combining these two sums yields kd = 2 E/=o 22t ~ Ei=o 22l+1 = Ei=o 22j_1i so
that d < kd, and Q) (k') is not a hyperoval in PG (2, 2/l).
When q > 2, a little more work is needed. Our treatment of this case
is adapted from similar arguments in Cherowitzo and Storme [5] dealing with
k = 2i + 2T
Lemma 3.10 Let k = 22 + 211 + 2*2 and let c\ (*i, h) and c\ (i2, h), where c > 1,
with io ^ 0 mod c. Now let i\ = i'2 ijf and h' Further, let k' =
2li + 2*2. If there exists a d! such that d! -< k'd' and such that if E d' = Ereh 2r
and 2l2 d' = E re/22r, h L\ I2 = $ (the products never coincide), then S> (k) is
not a hyperoval in PG (2,2h).
Proof: Where d' = EreD 2"> let d = EreD 2Cr Now since ^ld! = Ereh 2">
then 2ild = Ereh ^ Similarly, if 2^d' = Ereh^> then 2*ad = Ereh2^-
Since Ji fl I2 0 (the products never coincide), all terms remain distinct, and
34
since each r G D is realized as T for some r G /j U /2, each cr with r G D is
realized as 2cr for some r G I\ U /2.
Finally, since d = J2t w^ere r ^ 0
mod c. Thus, d -< kd, and (k) is not a hyperoval in PG (2,2h).
We are immediately able to require that (zi, h) (z2, h) c.
Lemma 3.11 Suppose k = 2l + 2n + 2l2; and let c\ (zl5 h) and c| (z2, h), where
(zi, h) { (z2, h), and where c > 1. Further, let zo ^ 0 mod c. Then S> (k) is not
a hyperoval in PG (2, 2^)
Proof: Let i[ = d-, z2 = and h! Then necessarily (i[,h') = ^ \
and \ (i'2,h') = Then the proof of Lemma 3.7 in [5] provides the
h'
value d! = 2a(tl,h ) for which all exponents in k'd' (as defined in Lemma
3.10) are distinct. Thus, this value of d' satisfies the conditions of Lemma 3.10
and @ (k) is not a hyperoval in PG (2, 2h).
The next lemma is an extremely close adaptation of Lemma 3.9 in [5].
Lemma 3.12 Suppose k = 2l + 2l1 + 212, and let c = ((zi,/i), (i2,h)), where
c > 1, and io ^ 0 mod c. Let i\ = ci'ly *2 = ci2, and h = ch!. If St (k) is a
hyperoval in PG (2, 2h), either i2 = 2i[ mod h' or i'2 = mod h', with h'
odd.
Proof: If c {h,h) or c / (z2,h), Lemma 3.11 applies and Q) (k) is not
a hyperoval in PG (2,2hy Then we must have c = (ii,h) = {i2,h), so that
(z^, h') = 1. Thus, we may write a = 2d and a1 = 2*2, and we may further write
h' = mi + j where m > 1 and j < i 1. In this case, the proofs of Lemmas
35
3.4 and 3.5 in [5] provide values d! = J2t=o at + a'+U and (if 3 = i 1),
d! = o at + oc2l~2 + YT=2 atl~l satisfying the conditions of Lemma 3.10 that
eliminate every case except z'2 = 2i[ mod h' and i'2 = mod h! with b!
odd.
We now consider the remaining possibilities. Since (z2, h!) = 1 as well, we
must, in addition to having either z2 = 2i[ mod h! or i2 = mod h' with
h' odd, have either i[ = 2i2 mod h' or i[ = mod h' with h! odd. Notice
in particular, that if i2 = i[^j^- mod h!, 2z2 = i[ (h' + 1) = i[ mod h'. As
such, we have two possibilities: either i\ = 2z2 mod h' with h! odd, or i\ = 2z'2
and i2 = 2i[.
The next two lemmas, closely adapted from Lemmas 3.11 and 3.12 in [5]
rule out these cases.
Lemma 3.13 Suppose k = 2* + 2n + 22n, where (ii, h) = (2*i, h) = c > and
(z0, c) = 1, with h! = ^ odd. Then Â£> (k) is not a hyperoval in PG (2, 2h).
Proof:
Let d = Yt=o 22tl1 Then the following is obtained for kd:
h'-l
2
h' + l
2
h'-l
2
l+*0
kd = E 22(t+l)ii _j_ E 22til + 22til
t=0 Z=1 t=0
h'-3 h'-l
2 2 _____________________
= 2h + ^2 22(m)il + 1 + 22til + 2h + 22til
t=l t=i
ft'-i
= 1 + 2il+1 + ^ 2th + ]P 22til+io
h'-l
2
t=0
+20
Z=2
t=0
Since Zi + 1 = 1 mod c, while each ti\ = 0 mod c, and since 2ti\ + z0 ^ 0
mod c the only terms which may possibly not be distinct are 2ll+1 and 22Ul+t.
But if c > 2, even a possible 2 2n+l =2 mod c, so that the resulting terms are
36
distinct and d < kd. On the other hand, if c = 2, Lemma 3.9 yields the desired
result, so that (k) is not a hyperoval in PG (2,2h).
Lemma 3.14 Let k = 2l+2n+2t2, with (i1; h) = (i2, h) = c > 1 and (i0, c) = 1.
Let h' = and i'2 = and suppose that i2 = 2i\ mod h! and that
= 2i2 mod h'. Then @ (k) is not a hyperoval in PG (2, 2h).
Proof: Since i2 = 2i[ mod h' and i[ = 2i2 mod h1, then z2 = 2ix mod h and
i\ = 2i2 mod h, so that {*i,*2} = {|, ^}. But then h' = 3, which is odd, and
Lemma 3.13 applies, so that (k) is not a hyperoval in PG (2, 2ft).
Theorem 3.15 Let k = 2l + 2n + 212, with (io,h) > 1 and(i\,h) = (z2,/i) > 1,
where i0 ^ 0 mod (zl7 h). Then (k) is not a hyperoval in PG (2, 2h).
Proof: The conditions stated satisfy the hypotheses of Lemmas 3.9, 3.10, 3.12,
3.13, and 3.14, which leave no possibilities for k to describe a hyperoval.
We note that, although the previous restrictions allowed us to assume that
(*i,h) = (i2,h), this was not assumed for the initial lemmas and was, in fact,
ruled out in Lemma 3.11, so that this series of lemmas in fact ruled out the
existence of any m dividing i\, *2, and h, but not i0.
With the added condition that k be irreducible, we can, in fact enforce a
further restriction.
Lemma 3.16 Suppose k = 2l + 2n +2*2 is irreducible and (i0, h) > 1, (i\,h) >
1, and (*2, h) > 1. Then |{(*0, h), (ii,h), (*2, h)}\ = 2 or @ (k) is not a hyperoval
in PG (2,2h).
37
Proof: Lemma 3.5 guarantees that |{(z0, h), {i\,h), (z2, h)}| ^ 1, so suppose
that |{(zo, h), (ii, h), (z2, h)}| = 3. Then Lemma 3.8 guarantees that z0 = i\ =
1 mod (z2,/z), i0 = i2 = 1 mod (z^/z), and zi = z2 = 1 mod (z0,h).
These conditions ensure that any prime dividing (z0, h) does not divide i\ or z2,
any prime dividing (z!, h) does not divide z0 or z2, and any prime dividing (z2, h)
does not divide z0 or i\. Thus, there are at least three distinct primes dividing h,
so that k is reducible, contradicting Theorem 3.4. So |{(z0, h), (zi, h), (z2, h)}\ =
2, as desired.
With these preliminary steps in place, we are now prepared to prove the
main theorem.
Theorem 3.17 Suppose k 2l + 2n + 212 is irreducible with (io,h) > 1,
(zi, h) > 1, and (z2, h) > 1. Then (k) is not a hyperoval in PG (2, 2h).
Proof: Lemma 3.16 ensures that under these conditions, exactly one of the
greatest common divisors of z0, ii, and z2 with h is distinct. Without loss of
generality, assume (z0, h) ^ (zi, h) (z2, h). Further, by Theorem 3.4, h = paqP.
Lemma 3.7 guarantees that i\ = z2 = 1 mod (io,h), so that (io,h) has no
common factors with i\ or z2. Thus, without loss of generality, (io,h) = pa and
(ii,h) = qb, where 1 < a < a and 1 < b < (5. Now let {z/,zm,zn} = {zo,ii,z2}
in some order. Since k is irreducible, potqli~l divides one of the differences,
without loss of generality in im. Further, pa~lq@ divides one of the differences,
which cannot be in im lest h divide in im. Without loss of generality, then
pa-iq0\^m ih So pQ_1^_1 divides both in im and im ii, and therefore also
divides Z/ in. Thus, pa~lqP~1 divides each of zq z2, i\ z0, and z2 i\.
38
Now define o' = min {a 1, o} and b' = min {(3 1, b}. So pa' divides each
of i0, *o *2, and i\ i0, implying that pa' also divides i\, i2, {ii, h), and (i2, h).
Since (fy h) = q13, a' = 0, so that a a = 1. Similarly, qb> divides each of i\, and
i\ i0, implying that qv also divides i0 and thus (iQ, h). Since (i0, h) = pa, b' = 0,
so that b = j3 1. Thus, h = pq, q divides both ii and i2, and i\ = i2 = 1
mod p. Since the p multiples of q have distinct congruences modulo p, ii = i2,
a contradiction. So (k) is not a hyperoval in PG (2, 2h). m
Thus, any monomial hyperoval in PG (2,2h) with irreducible exponent k =
2l + 211 + 2t2 must satisfy (fy h) = 1 for some j.
Note, however, that the sequence |(21>)I| is merely a permutation of the
sequence {2*}^. If then we let a = 2fy we can express each integer in Z2h_!
in the form n = q ai(xl, where Oj {0,1} and the summation is reduced
modulo 2h 1 to lie in [0, 2h l] with the convention that 0 is reduced to 0
and any nonzero multiple of 2h 1 to 2h 1. Further, the a* are simply a
permuatation of the bits in the binary expression of n, and this permuation
is invariant under the choice of n. Defining the same partial ordering for this
expansion as in Definition 2.25, it is immediately evident that a -< b in the
binary expansion if and only if a ^ b in the expansion with respect to a. It
follows then, that Theorem 2.26 applies equally well to this expansion. We may
thereby restrict our attention to those exponents of the form a + a1 + cF, and,
in so doing, reduce the number of parameters to be considered by one.
Although Glynns criterion applies to this expansion, there is an added dif-
ficulty when terms coincide. In the binary expansion, the term 2 2a is easily
simplified to 2a+1. In the a-ary expansion, the simplification depends completely
39
on the choice of a. In a sense, then, we have not eliminated a parameter, but
have simply changed its nature. Our work with the a-ary expansion will often
require the parameter r, where ar = 2, to simplify any terms which coincide as
2 a = aa+r.
3.3 Monomials With Exponent of the Form a + a1 + ai
It is convenient to consider the relationships between h, j, and i. Without
loss of generality, we will assume j > i. We assume throughout that a = 2lh
where {ij,h) = 1. Using the division algorithm twice, we may express h as
h = mj + ni + l, where ni + l < j and l < i. We are immediately able to
exclude all but a few monomials from the set of monomial hyperovals. Although
reducibility of the exponent guarantees that the exponent have this form to be
a hyperoval exponent, we do not here assume that the chosen values of k be
reducible, and the results that follow are true for any value of k that can be
expressed as a + a1 + a].
Lemma 3.18 Let k = ot+a%+a?, where a = 2l for (i0, h) = 1 and suppose that
for some s > 1, s|i, s\j, and s|h. Then @ (k) is not a hyperoval in PG (2, 2h).
Proof: Without loss of generality, suppose that a1 = 2n and ad = 212. Then
s|ii and s|2, but s \ *o, and the results of Lemma 3.11 and Theorem 3.15
preclude Q) (k) from being a hyperoval in PG (2,2h).
Our next result shows that, in fact, once k can be expressed as a + a1 + af
one of i and j must be relatively prime to h, and therefore k can also be expressed
as (3 + Px + /3y, where fdx = a and (3 = a1 if (z, h) = 1 or (3 a.i if (j, h) = 1.
40
Theorem 3.19 Let k = a + a1 + a3, where a = 2J and aT = 2. Suppose that
(i, h) = a and (j, h) = b for a > 1 and b > 1. Then & (k) is not a hyperoval in
PG( 2,2h).
Proof: By Lemma 3.18, we may assume that (a, b) = 1, so that ab\h. If
i^l mod b, let d = ^2b=o a<:bi so that kd = J2c=o (c6+1 + a.cb+1 + acb+3).
Since b\j, Yhc=o Oicb+3 = Y2c= o aCb = d. No terms coincide as i ^ 0 mod b
(since b \ z)and z ^ 1 mod b. On the other hand, if i = 1 mod b and r ^
1 mod 6, kd \aCb + otcb+r+1) and no terms coincide, so that d ^ kd
__^ h_2
in either case. Similarly, if j ^ 1 mod a, let d = X^c=o aCa> so that kd =
X^c=o + ca+1 + a004"-7)- Again, no terms coincide as j ^ 0 mod a (since
a f j) and j ^ 1 mod 6. Once again, if j = 1 mod a and r ^ 1 mod a,
fe _2
kd = Yhc=o (QCa + aca+r+1) and no terms coincide, so that d < kd in either case.
Thus, the only possibility is that i = 1 mod b, j = 1 mod a, and r = 1
mod ab. Notice then that i+j = 1 mod ab, since i+j = 1 mod a and i+j = 1
mod b. Notice further that j ^ 2i, as a \ j, and that i > 2 (and consequently
j > 3), as b > 1.
If now j 2i 1, we choose d = (CQb + acab+l_1), so that kd =
Ecto1 (ac6 + ca6+1 + 2acaf,+i + acab+2i-1+aca6+J). With r = -1 mod ab,
ZÂ£01 2acab+t = ]P=o1 acab+t_1 and no other terms coincide, hence d + kd.
On the other hand, if j 2i, let d = q1 (ctcab + acab+3~1), so that kd =
o1 (acab + acab+1 + acab+1 + acab+3~l + 2o;cab+-?). Again, r = 1 mod ab, so
that J^c=o 2aC(lb+] = f2c=o oaib+3'"1 and no other terms coincide, so again
d + kd.
Thus, having covered all possible cases, if (k) is a hyperoval in PG (2, 271),
41
either (i,h) 1 or (j, h) = 1.
We now use five related values of d to place severe restrictions on the param-
eters of a potential hyperoval. After each, we consider the restrictions on the
parameters resulting from combining the restrictions from all of the values of d
used up to that point. This results in a new set of restrictions on the parameters
that is carried forward and combines the information in all of the values of d
used to that point.
Lemma 3.20 Let k = a + al -\-ai, and let h = mj + ni + l, where ni + l < j and
l < i. If *3) (k) is a hyperoval in PG (2,2h}, then at least one of the following
conditions holds:
1. i-l = l
2. j i
3. j 1 = ni + l.
Proof: Suppose that none of the above conditions hold, and let d = Y^!a=o a<1 +
Eteo'a^' + KTo1 acj+m+i Thjg ieacis to the following value for kd.
I-1 711
(oa+1 + aa+i + aa+j
a=0 6=0
m 1
_j_ ^acj+ni+l+1 acj+ni+l+i _|_ Qcj+ni+l+j
c=0
Under the conditions set out, d is defined. Further, all terms of d are present and
no terms coincide, so that d < kd and & (k) is not a hyperoval in PG (2, 2h).
) + >>
+ a(
bi+l+i
+ '
bi+l+j'j
42
Lemma 3.21 Let k = a + al + aj, and let h mj + ni + l, where ni + l < j and
l < i. If (k) is a hyperoval in PG (2,2h), then at least one of the following
conditions holds:
1. n < 1
2. i = 2
3. rn = 1 and j = ni + l + 1
4- m = 1 and j = ni + l + i 1
5. m = 1 and j = ni + l + i
6. 1 = 0.
Proof: Suppose that none of the above conditions hold, and let d = Y?a=o a<1 +
oP+i-i + ai+m+l~l + ac^+TU+b We then obtain the following
expression for kd.
i-i
n1
+ aj+ni+l~i+1 -p (yi+ni+l a2j+ni+l-i
m 1
+
Again, under the conditions imposed, d is defined, all terms of d are present,
and no duplicate terms appear. Thus, d < kd and (k) is not a hyperoval in
Corollary 3.22 Let k = a + a1 + aJ, and let h = mj + ni + 1, where ni + l < j
and l < i. If (k) is a hyperoval in PG (2,2hthen at least one of the following
conditions holds:
1. n = 1 and i 1 = l
2. n = 0 and i 1 = l
3. i 2 and l = 1
4 n = 1 and j i*
5. n = 0 and j i**
6. m = 1 and j = ni + Z + i 1*
7. m = 1 and j = ni + l + i
8. j = i + 1+ 1 and n = 1
P. m = 1 and j = ni + l + 1
.70. j = m + 1.
Proof: We simply check that anything satisfying each set of conditions from
Lemma 3.21 must also satisfy one of the conditions above. Notice that if S (k)
is a hyperoval some set of conditions from Lemma3.20 must also be satisfied.
If n > 1 and i 1 = l, either condition 1 or condition 2 holds. If n > 1 and
j i 1 and
j l ni + l, then n = 1 and condition 10 holds.
44
If % 2 and i l = l, condition 3 holds. If i = 2 and j i < ni + l < j i + l,
then either l = 1 and condition 3 again holds, or l = 0 and thus j = ni+i so that
2| (i, h) and 2| (j, h), a contradiction of Theorem 3.19. If i = 2 and j 1 = ni + l,
either l = 1 and condition 3 holds, or l = 0 and condition 10 holds.
Conditions 3, 4, and 5 of Lemma 3.21 are identical to conditions 9, 6, and
7 respectively.
If l = 0, % 1 ^ l as i > 1. If / = 0 and j i
and i divides each of i, j, and h, a contradiction of Theorem 3.19. Finally, if
l = 0 and j 1 = ni + l, condition 10 holds.
Lemma 3.23 Let k = a + a1 + cF, and let h = mj + ni + l, where ni + l < j and
l < i. If (k) is a hyperoval in PG (2, 2h), then at least one of the following
conditions holds:
1. n 0
2. i = 2
3. j = ni + l + 1
4. j i
5. j 2 i
6. 1 = 0.
Proof: Suppose that none of the above conditions hold and let d = a
0,1+i-i _|_ abl+l + Y1T=() Q'CJ+m+/. We then obtain the following expression
45
for kd.
i-1
kd = J2 (Q+1 + a+i + a+j) + i+l + 2i+'_1 +
a=0
n1
+ ^ (<*W+,+1 + abi+l+l + aj+bi+l)
6=1
m1
^cj+m+z+l acj+m+i+; _|_ Qcj+j+ni+l^
c=0
Again, under the conditions imposed, d is defined, all terms of d appear, and
no duplicate terms appear. Thus, d A kd and (k) is not a hyperoval in
PG( 2,2h). u
Corollary 3.24 Let k = a + a* + ckj, and let h mj + ni + 1, where ni + l < j
and l < i. If & (k) is a hyperoval in PG (2, 2h), then at least one of the following
conditions holds:
1. n = 1, i \ l, and j = 2i
2. n = 0 and i 1 = l
3. i = 2 and l = 1
4- n = 1 and j i *
5. m = 1 and j = ni + l + i 1*
6. m = n = l = 1 and j = 2i
7. m = 1, j = ni + i 1 and l = 0
5. m 1 and j = ni + l + i
46
9. n = 1 and j = i + l + 1
10. m = 1 and j = ni + l + 1
j = ni + 1 and l = 0
12. n = 0 and j < i + l
Proof: Conditions 2, 3, 4, 5, 6, 7, 8, and 9 of Corollary 3.22 are identical to
conditions 2, 3, 4, 8, 5, 12, 9, and 10 respectively.
Condition 1 of Corollary 3.22 contradicts conditions 1 and 6 of Lemma 3.23,
satisfies condition 3 whenever it satisfies condition 3 of Lemma 3.23, satisfies
condition 1 whenever it satisfies either of conditions 3 or 5 of Lemma 3.23, and
satisfies condition 4 whenever it satisfies condition 4 of Lemma 3.23.
Condition 10 of Corollary 3.22 contradicts conditions 1, 4, and 5 of Lemma
3.23, satisfies condition 3 or 11 whenever it satisfies condition 2 of Lemma 3.23,
and satisfies condition 11 whenever it satisfies either of conditions 3 of 6 of
Lemma 3.23.
Lemma 3.25 Let k a + cd + ed, and let h = mj + ni + l, where ni + l < j and
l < i. If Q) (k) is a hyperoval in PG (2,2^), then at least one of the following
conditions holds:
1. l = i- 1
2. j = 2i
3. j = i + 1
4- m = 1 and j = ni + l + 1
47
5. m = 1 and j = ni + i + l
6. n = 0
7. m = 1, n = 1 and 2i < j < 2i + l
Proof: Suppose that none of the above conditions hold and let d = X^L=o a +
Eto1 ^+ + QI+"-'+' + E27 acj+m+i' yye obtain the following expression for
kd:
i-1
711
*<*=Â£(
oa+1 + aa+i
+ aa+j) + Y (aW+m + a+ aj+bi+l)
a=0
6=0
_|_ aj+ni-i+l+1 _|_ aj+ni+l _|_ ^2j+ni-i+l
m 1
+ (acj+ni+l+1 +
a
cj+ni+i+l _|_ acj+j+ni+l
')
c= 1
Again, under the conditions, d is defined, d appears in the expansion of kd, and
no duplicate terms appear in the expansion of kd so that d ^ kd and @ (k) is
not a hyperoval in PG (2,2h).
Corollary 3.26 Let k = alpha + cd + a?, and let h = mj + ni + l, where
ni + l < j and l < i. If Â£> (k) is a hyperoval in PG (2, 2h), then at least one of
the following conditions holds:
1. n = 0 and i 1 = l
2. i 2 and l = 1
3. n = 1, i 1 = l, and 2i < j < 3i 1
4- m = 1, j = ni + 2i 2, and i 1 = l
48
5. n = 1 and j = 2i
6. n 1, l = 0, and j = i + 1
7. m = 1 and j = ni + l + 1
5. m = 1 and j = ni + i + l
9. m = 1, n = 0; and j = z + l 1
m = n 1 and 2i < j <2i + l
11. j < i + l and n = 0.
Proof: Conditions 1, 2, 6, 8, and 10 of Corollary 3.24 and conditions 4, 5, and
7 of Lemma 3.25 imply conditions 3, 1, 10, 8, 7, 7, 8, and 10 respectively.
Condition 1 of Lemma 3.25 satisfies condition 3 whenever it satisfies either
of conditions 4 or 9 of Corollary 3.24, satisfies condition 4 whenever it satisfies
condition 5 of Corollary 3.24, and contradicts conditions 7 and 11 of Corollary
3.24.
Condition 2 of Lemma 3.25 satisfies condition 5 whenever it satisfies either
of conditions 4 and 9 of Corollary 3.24 and contradicts conditions 5, 7, and 11
of Corollary 3.24.
Condition 3 of Lemma 3.25 satisfies condition 6 whenever it satisfies any of
conditions 7, 9, and 10 oc Corollary 3.24, and contradicts conditions 4 and 5 of
Corollary 3.24.
Condition 6 of Lemma 3.25 satisfies condition 9 whenever it satisfies condi-
tion 5 of Corollary 5 and contradicts conditions 4, 7, 9, and 11 of Corollary 5.
49
Lemma 3.27 Let k = a + al + aj, and let h = mj + ni + l, where ni + l < j and
l < i. If 31 (k) is a hyperoval in PG (2, 2h), then at least one of the following
conditions holds:
l.i = 2
2. j = i + 1
3. ni + i < j < ni + i + l 1
4- m = 1
5. 1 = 0.
Proof: Suppose, as usual, that none of the above conditions hold, and let
d = aa+Sb=i of1*11+aj+ni+l~1^ aci+ni+l. We obtain the following
expression for kd:
i-1
kd
=E(
att+1 + aa+i + aa+j
')+Â£(
abi+l + a6i+i+i-l
+ a'
bi+j+l-
)
a=0
6=1
_|_ cyj+ni+l _|_ aj+ni+i+l-1 a2j+ni+l-l
m1
acj+ni+l+1 acj+ni+i+l _|_ cj+j+ni+l^
C=1
As before, d is defined, the terms of d appear, and no duplicate terms appear,
so that d -< kd and 3! (k) is not a hyperoval in PG (2, 2h).
Our final combination of conditions results in eleven cases. Because this
provides the structure for much of the work which follows, we state this as a
separate theorem rather than a corollary.
50
Theorem 3.28 Let k = a + a1 + aj, where a = 2io and (io,h) = 1. Further
let h = mj + ni + l, where ni + l < j and l < i. If St (k) is a hyperoval in
PG (2, 2h), at least one of the following sets of conditions holds:
1. n = 1, m = 1, and 2i + l
2. i = 2 and l = 1
3. m = 1, j = i + l 1, and n 0
4- i 1 = l, m = 1, and j ni + 2i 2
5. j = ni + l + 1 and m = 1.
6. n = 1, j = 2i, and l ^ 0
7. m = 1 and j = ni + i + l
8. n = 1, j = i + 1, and l = 0
9. i 1 = l, n = 1, and 2i < j < 3i 2
10. n 0 and j < i + l
11. i 1 = l, n = 0, and m = 1
Proof: Conditions 2, 4, 5, 6, 7, 8, 9, 10, and 11 of Corollary 3.26 fall into
cases 2, 4, 1 (since if l 0, i divides each of i, j, and h, contradicting Theorem
3.19), 8, 5, 7, 3, 1, and 10, leaving only conditions 1 and 3 of Corollary 3.26 to
be considered.
Condition 1 of Corollary 3.26 falls into case 2 whenever it satisfies condition
1 of Lemma 3.27, falls into case 10 whenever it satisfies conditions 2 or 3 of
51
Lemma 3.27, falls into case 11 whenever it satisfies condition 4 of Lemma 3.27,
and contradicts condition 5 of Lemma 3.27.
Finally, condition 3 falls into case 2 whenever it satisfies condition 1 of
Lemma 3.27, contradicts conditions 2 and 5 of Lemma 3.27, falls into case 9
whenever it satisfies condition 3 of Lemma 3.27, and falls into case 1 whenever
it satisfies condition 4 of Lemma 3.27.
We examine each of these cases in turn as the major portion of this classi-
fication, obtaining the following theorem. As the proof consists of a very long
case argument that follows the pattern of the preceding lemmas and is not par-
ticularly instructive, it is left to the next chapter.
Theorem 3.29 Let k = a + a1 + a?, where a = 2l and (z0, h) = 0. Let
h = mj + ni + l, where ni + l < j and l < i. If & (k) is a hyperoval in PG (2,2h)
it is a translation hyperoval, Segre hyperoval, or Glynn hyperoval.
3.4 Reducible Monomial Hyperovals
Having completed the classification of monomial hyperovals with irreducible
exponents and those of the form k = a + a1 + af where a = 2l with (i0, h) =
1, we now turn our attention to the remaining case: those having reducible
exponents which cannot be expressed asa + a' + a^ under the restrictions on a.
We thus consider k = 2l + 2n + 212. The results of Lemmas 3.6 and 3.7
force (z0, h), (A, h), and (z2, h) to be distinct, and thus i\=ii = 1 mod (z0, h),
iQ = z2 = 1 mod (ii, h) and z0 = Zi = 1 mod (z2, h). With the appropriate
reduction and a reference to the classification result obtained for those hyper-
ovals of the form St {a + a1 + cF) with a = 210 and (zq, h) = 1, we are able to
52
rule out this final case as well.
Theorem 3.30 If (i0,h), (i\,h), and (i2,h) are distinct and each greater than
one, Â£> (k) is not a hyperoval m PG (2,2h).
Proof: By Lemma 3.8, i\ = i2 = 1 mod (z0,fi), i0 = i2 = 1 mod (i\,h),
and i0 = i\ = 1 mod (i2,h). It follows that (i0,h), and (i2,h) are
pairwise relatively prime. So (i0,h) (i\,h) divides h.
Now, since (io,h) (i\,h) is the least common multiple of (i0, h) and
*2 = 1 mod (i0, h) (*!, h). Further, i0 is congruent mod (*0, h) (ij, h) to
a multiple of (i0,h) and is congruent mod (i0,h)(ii,h) to a multiple of
(i\,h). Since, again, the product (z0,/i)(ii,fi) is the least common multi-
ple of the two factors, neither of which is one, 0, *i, and i2 have distinct
residues mod (i0, h) (ii, h). In particular, a ^-reduction of k exists with
k' = 2a<'l0,h) -f 2b(ll'h'> + 2(o.h)(*iA)-1, where 1 < a < (ii, h) and 1 < h < (io, h).
In such a reduction, neither a(i0,h) nor b(i\,h) can be relatively prime to
(io,hfc>r However> since 1 necessarily is relatively prime to {ioA^h),
this reduction describes a monomial hyperoval if and only if it is one of the
monomial hyperovals determined in Theorem 3.29. Since each such monomial
hyperoval of the form 2l + 2-y + 2h~1 has at least one of i or j relatively prime
to h, the reduction is not a hyperoval, and thus, by Theorem 3.3, 3 (k) is not a
hyperoval in PG (2, 2h).
Theorem 3.31 Let k = 2l + 2n + 212, with (i0, h), (ix, h), and (i2, h) all greater
than one. Then @ (k) is not a hyperoval in PG (2, 2h).
Proof: This is an immediate consequence of Lemma 3.5, Lemma 3.6, Lemma
53
3.8, Theorem 3.30, and Theorem 3.15.
We finally state the major theorem resulting from all of this work.
Theorem 3.32 If Qt (2l + 2n + 212) is a hyperoval in PG (2, 2h), it is a trans-
lation hyperoval, Segre hyperoval, or Glynn hyperoval.
Proof: This follows immediately from Theorem 3.29 which classifies those with
(ix, h) = 1 for some x {0,1,2} as the translation hyperovals, Segre hyperovals,
and Glynn hyperovals, and from Theorem 3.31 which rules out the possibility
of such a hyperoval with (ix, h) > 1 for all r G {0,1,2}.
54
4. Proof of Theorem 3.29
In this chapter, we exhibit a complete proof of Theorem 3.29.
Theorem 3.29 Let k = a + a1 + a?, where a 2l and (io,h) = 0. Let
h = mj + ni + l, where ni + l < j and l < i. If (k) is a hyperoval in PG (2,2^)
it is a translation hyperoval, Segre hyperoval, or Glynn hyperoval.
Theorem 3.28 places restrictions on the relationships among the parameters,
dividing these parameters into eleven distinct cases. We examine each of these
cases in turn, establishing for each a lemma requiring that any hyperoval satis-
fying the assumptions of that case be a translation hyperoval, Segre hyperoval,
or Glynn hyperoval.
4.1 Case 1
We turn now to case 1, in which n 1, m = 1, and 2i + 1 < j < 2i + l. For
ease of notation, let j 2i + x, where 1 < x < l. Further, let r be the unique
power such that ar = 2, with 1 < r < h 1. We use the family {da} of values
for d, where
la i+l
da = 'Y2ab + ab
6=0 b=i+la
and 0 < a < l.
In this case we obtain the following product kda:
la+l i+l+1 2 i+l 2 i+x+la Si+x+l1
kda ^ ab + ^ ab + ^ ac + ^ ac + ^ ab.
6=0 b=i c=2i+la c=2i+x b=3i+x+la
Notice that within this product, the only terms which may possibly coincide
are those indexed by c, unless l = i 1 and a = 0 or a = l. Thus, the
55
powers of a that appear multiple times are precisely the powers ac, where
max {2i + x, 2i + l a} < c < min {2i + l + x a, 2i + /}. Notice also that
all terms of da appear in kda. The use of {da} severely restricts values which
may be taken by r.
Proposition 4.1 Let k = a + a1 + a2l+x, where a = 2l with aT = 2 and
1 < r < h l. Further, let h = 3i + l + x, where l < i and 1 < x < l. If
min {x,l x} + l
3i + l 1, orifl^i 1 and r E {i + l + 2, 2i + l 1,2i + l + x + 2, i + x 1},
then (k) is not a hyperoval in PG (2, 2h).
Proof: Given a value of a, we determine the values for r for which otc+r does
not appear for any max {2i + x,2i + l a} < c < min {2i + l + x a,2i + 1}
and thus da X kda. We consider each of the four possible combinations for
the maximum and minimum terms separately. If 2i + x > 2i + l a and
2i +1 + x a > 2i +1, the acceptable values of r are i + 2l a + 2 < r < 2i + x 1,
and a satisfies l x < a < x. If2i + x>2i + l a and 2i + l > 2i + l + x a, the
acceptable values for r are i+21a+2
and lx+1 < r < i1, and a satisfies max {x,l x} < a < l. If 2i+la > 2i+x
and 2i + l + x a > 2i + l, the acceptable values for r are i + l + x + 2 > 2i + x l,
2i + l + x + a + 2 < r < 3i + 2x 1, and x + 1 < r < i + x a 1, and a satisfies
a < min {x, l x}. Finally, if 2i + l a > 2i + x and 2i + l > 2i + l + x a, the
acceptable values of r are i+l+x+2
and a + l
Considering at once all possible values of a in a given situation, the values
i + l + 3
56
x-f3
min {x,l x} + 1 < r < i + x 2 (min {x, l x}-\-l
each have some choice of a with 1 < a < Z 1 (0 < a < Z if l < i 1) such that
da zf: kda, yielding the desired result.
Proposition 4.2 Let k = a + a1 + a2l+x, where a = 210, ar = 2, and h =
3i + l + x, where l < i and 1 < x < l. If r < min {x, l x} or r > 3z + l, (k)
is not a hyperoval in PG (2, 2h) or it is a translation hyperoval, Segre hyperoval,
or Glynn hyperoval.
Proof: Consider the value kdj. Notice that kdi has only the duplicate terms
2a2i+i-i an(j 2a2l+l and no terms ac with 2i + x + l a + 1
appear. Since this amounts to i consecutive terms and i > min (x, l x}, there
exist q, q such that a2l+l~1+qT and a2l+l+q'r do not otherwise appear in kd\,
while for s < q, s' < q', a2l+l~l+sr and a2l+t+s'r never appear in d\, so that
d\ < kd\.
In similar fashion, notice that r > Si + Z is equivalent to h r < x. No
terms ac with i + l + 2
i + x l 2>x consecutive terms, except when Z = i 1 and x > i 2. Thus,
unless Z = i 1 and x > i 2, using the same reasoning as above, cZi < kd\.
Now if Z = i 1 and x = i 2, the only duplicate terms are 2o?l~2 and 2a3l_1,
while if Z = x = z 1, the only duplicate term is 2a3,_1, and neither aM~2+r nor
q3ji+r appears otherwise for r > 3% + 1 + 1. Thus, we have only to consider
r = Si + Z in these two situations.
57
If Z = i l=x+l = h r 1, assume i > 3 (if i = 2, Lemma 4.21 yields
the result), and let d = 1 + cd_1 + o?l~x + a3*-2, so that kd 1 + a + ed-1 +
2cd 4- 2l_1 + a21 + a3l~2 + 2a3*-1. As 2cd = a2 and 2a31'1 = o:2l+1, neither of
which terms appear otherwise, d < kd.
If Z = z = d r = i 1, let d = 1 + Y^c=i\i aC> so that = 1 + a + cd +
Sc=^+2 a + 2o;3l_1 + Y^c=Ai a- Since 2a3*-1 = a2t and 2a21 = cd+1, d < kd, so
that the result follows.
The preceding propositions now restrict the values taken by r in this case
toi + r
i + x 1 < r < i +1 + 2 and 2i + l 1
examine these remaining cases, beginning with 2i +1 1 < r < 2i +Z + x + 2. In
this case we are immediately able to show that r is restricted to the boundaries
of this range.
Proposition 4.3 Let k = a + cd + a2l+x, where a = 2l with cd = 2 and
1 < r < h l. Further, let h = 3z + l + x, where l < i and 1 < x < l. If
2i + l
Proof: Choose d = YX=o6 + a3l+l+x~r~1. Since r < 2i + x + l, all of these
terms are distinct. Then kd = Y2l=.v ab + a>> + a3l+l+x~r + a4l+l+x~r~l +
a5i+i+2x-r-i' gjnce r > 2i + l > i + l + x + 1, Qpt+t+x-r-1 appears in kd,
and a3l+l+x~r appears exactly twice unless r = 2i + l and l x = i 1.
But 2a3l+l+x~r = 1, and if no further powers of a coincide, d < kd. Since
r > 2i + l > 2i + x, 5i + l + 2x r 1 < 3i + l + x, and a5l+l+2x~1-1 does not
coincide with any other terms. Finally, since r > 2i + l, 4i + l + x r 1 < 2i + x,
58
so that a4l+l+x~r~1 does not coincide with any other terms. Thus, d ^ kd, and
(k) is not a hyperoval.
In the case r = 2i + l and l x = i 1, let d = Ec=o a + so that
kd = I^cIoac+2cd+X^=i+i ac+2a3*'~1 + E^='3i ftC- Now 2cd+2a3_1 = cd+a:4i,
and d < kd again.
This leaves r = 2i + l 1 (when l = i 1), r = 2? + / + x + 1, and
r = 2i + l + x + 2 (when l = i 1) among the higher set of remaining values for
r.
Proposition 4.4 Let k = a + cd + a2l+x, where a = 2l with 0i2l+l+x+l = 2.
Further, let h = 3i + l + x, where l < i and 1 < x < l. Then ^ (k) is not a
hyperoval in PG (2,2h).
Proof: If l < i 2, let d = E^o0^ + cd+* (previously denoted by do).
Then kd = ES + ElHi b + EjS+iab + ai+l+1 + <*2l+l + a3i+Z+X- Note
that a3l+l+x = 1 and that the only power of a that appears twice is a2l+l.
But 2a2l+l = a1+l+1. Since this already appears, we realize that 2o1+/+1 = al+2,
which cannot appear twice in the product kd. Thus, since all powers of a present
in d are present in kd, d < kd and L2) (k) is not a hyperoval in PG (2, 2h).
On the other hand, if l > i 2, let d = Eb^oab + ct2l_3- Then kd =
ab + 2a2t_3 + Eb^h+x ab + ol3i~3 + ait+x~3. It may happen that a 2o3t~3
term is present. But this is equal to cc2l~2 and we have either a 2a2*-2 or 3a2l~2
term present. In either case, this results in a second cd_1 term, which simplifies
to a 1. Thus, d < kd.
Proposition 4.5 Let k = a + cd + a2l+x, where a = 2l with a2l+l+x+2 = 2.
59
Further, let h = 3i + l + x, where l = i 1 and 1 < x < l. Then Q) (k) is not a
hyperoval in PG (2,2h).
Proof: Let d = 1 + Yl[=2+ al+l Then kd = a + a1 + a2l+x + ^6=3a>) +
ab + Y^b=2i+l+2ab + a21 + a2t+l + a3l+l+x. It may be that a 2a2l+l term
is present. But this is simply a2l+1, which does not otherwise appear in the
product, unless x = 1, in which case 2a2l+1 = cd+1, which does not appear.
In addition, a 2al term is present, which simplifies to a2, which is not
otherwise present. Thus, all powers of a present in d are also present in kd, so
that ^ (k) is not a hyperoval in PG (2, 2/l).
Proposition 4.6 Let k a + a1 + a2l+x, where a = 2l with a2lJrl~l = 2.
Further, let h = 3i + l + x, where l i 1 and 1 < x < l. Then @ (k) is not a
hyperoval in PG (2, 2h) unless it is a Segre hyperoval.
Proof: Consider di = 1 + ab as defied earlier. Then kdi = ]T^=0 ab +
J2b=i ab + Oi2l+x + a + q6- The only powers that appear more
than once are 2a21 and 2a2l+x. Now 2a21 = a4l+i_1 = a3t+l+l = al~x, and
so long as l x > 2, no further terms coincide. On the other hand, 2a2l+x =
a4i+i+x-1 which does not already appear. Since all terms of d are present
in kd, d
When l x = 1, k = 2 + 2h~3 + 2h~3. For > 3, let d = YLc=oa<:> so that
kd = &c + &h~5 + ah~4 + 2ah~3 + ah~2 + a^1. As 2ah~3 + ah~2 + a^1 = 1,
d A kd. If % = 2, k = 2 + 22 + 24 and h = 7, so that ^ = 6 and (k) is a
Segre hyperoval.
60
Finally, when l = x, notice that if i is even, 2| (3i 2, 5i 2). Hence i is
odd. We consider i = 1 mod 4 and i = 3 mod 4.
If i = 4s + 1, it is clear that h = 20s + 3. Further, it is easy to verify that
a = 215s+1, a1 = 210s+1, and a3l_1 = 215s+2. In this case, set d = l + 25s + 210s+1,
so that kd = 25'-1 + 25s + 210s+1 + 2 215s+1 + 215s+2 + 220s+1 + 2 220s+2, which
simplifies to 1 + 25s_1 + 25s + 210s+1 + 215s+3 + 220s+1, so that d ^ kd.
If i = 4s + 3, it is clear that h = 20s + 13. Further, it is easy to verify that
a = 25s+2, a1 = 210s+6, and a3l~l = 25s+3. In this case, set d = l + 25s+3 + 210s+7,
so that kd = 1 + 25s+2 + 25s+3 + 210s+5 + 2 210s+6 + 2 215s+9 + 215m+1, which
simplifies to 1 + 25s+2 + 25s+3 + 210s+7 + 215s+u, so that d ^ kd. m
We now address the situation in which i + x 1 < r < i + l + 2. We are again
able to immediately show that r is restricted to the boundaries of this range.
Proposition 4.7 Let k = a + a1 + a2l+x, where a 2l with ar = 2. Further,
let h = 3i + l + x, where l < i and 1 < x < l. Ifi + x
not a hyperoval in PG (2,2h).
Proof: Consider d = J2b^oX~r ab + a2l+(+x_r. Then kd = ^^1+x_r+1 ab +
ES<+Ir+i ab + ab + a3i+l+x~r + a4i+l+2x~r. Notice that under the
restrictions, 4i + l + 2x r < 3i + l + x and the only power of a that appears
twice is ocil+l+x-r. But 2a3t+l+x~r = a3l+l+x = 1, which does not otherwise
occur. Thus, d -
This leaves r = i + x, r = i + l + 1, r = i + l + 2 (when l = i 1), and
r = i + x 1 (when l = i 1).
Proposition 4.8 Let k = a = a1 + a2l+x, where a = 210 with at+x 2. If
61
h = 3i + l + x, where l < i and 1 < x < l, then 2> (k) is not a hyperoval in
PG(2, 2h).
Proof: Let d = 0^ + cd+/_x_1. Then kd = Y^b=i a& + &l+l~x + ot2l+l~x +
Y^b=2%+x ab + Oi3l+l. Now 2a'l+l~x = a2l+l appears, yielding 2a2l+l = 1. It is
possible that 2a2l+l~x = a3l+l appears. In this case, 2a3l+l = a\ 2al = a2l+x,
and 2oi2i+x = a3t+2x, which does not otherwise appear unless l = x, in which
case, a second a2l+x no longer appears. Thus, d X kd. m
Proposition 4.9 Let k = a + a1 + a2l+x, where a = 2l with al+x_1 = 2.
Further, let h = 3i + l + x, where l < i and 1 < x < l. Then 2 (k) is not a
hyperoval in PG (2,2h).
Proof: Setd=Â£'lx+V + a2l+'. Then kd = <*b + ^ +
Yll^i+x ab + 2a2l+,+1 + a3l+l. We notice that the only doubled powers of a are
2a1 (or 3a1 when x 1) and 2a2l+l+1. Now 2a1 = a2l+x~l, which does not occur
elsewhere in the product. Further, 2a2l+l+1 = o:3l+i+x=1, So that d -< kd and
2 (k) is not a hyperoval in PG (2, 2^).
Proposition 4.10 Let k = a + cd + Oi2l+x, where a = 210 with cd+*+1 = 2.
Further, let h = 3i + l + x, where l < i 2 and 1 < x < l. Then 2 (k) is not a
hyperoval in PG (2, 2h).
Proof: Consider di = 1 + ab Then kd = a + a1 + a2l+x + E&ti+i ab +
Y^b=2iab + Eiltl+xSince a3l+l+x = 1, all terms of d are present in the
product. In addition, a 2oi2i+x appears. But 2q;2j+x = a, and 2a = a1+l+2.
Since l < i 2, cd+i+2 does not otherwise appear in the product, and thus
62
d -< kd.
Proposition 4.11 Let k = a + a1 + a2t+x, where a = 2l with cP+i+1 = 2.
Further, let h = 3i + l + x, where l = i 2 and 1 < x < l. Then Lfl (k) is not a
hyperoval in PG (2, 2h).
Proof: Set d = Et=o ocb+ al+x~l+ a2l+2x~2. Then kd = EEi o^ + Ejii""1 ab +
ESS1 ab + ai+x + + a3i+2x~1 + a2i+2x^ + a3i+2x~2 + a4i+3x~2. Now
a4t+3x2 a2x jt js possible that a 22x term exists, however, no 2x can be
present in d. If a 2a:2x term exists, it simplifies (since l = i 2) to a2t+2x~1.
As such, either a 3a2l+2x_1 term or a 2a2l+2x_1 term exists. In either case, an
additional ax term is produced. But 2ctx = a2l+x~l, and the resulting 2a2l+x_1
term reduces to 1. All powers of a present in d being retained, d < kd and 2 (k)
is not a hyperoval in PG (2,2h).
Where l = i 1 rather than i 2, a slight adjustment to the value of d
eliminates most of the remaining possibilities.
Proposition 4.12 Let k = a + a1 + a2l+x, where a = 2l with al+l+1 = 2.
Further, let h = 3i + l + x, where l = i 1 and 1 < x < l. Then & (k) is not a
hyperoval in PG (2, 2h).
Proof: We consider three cases on the values of x. If x > 3, we consider
d = EIZl + ES-J + a2l+2*-3- In this case, kd = Â£Â£ + ElS +
a21-2 + ESS-2 ab + 2o2i+2l_2 + ESS-3 ab- is possible that a 2q;2x-2 =
q,2z+2x-2 ^erm exists, so either a 3a2l+2x~2 term or a 2a2t+2x~2 term exists. In
either case, an additional ax_1 term is produced. But 2ax_1 = a2l+x~4, and the
63
resulting 2a2l+x 1 term reduces to 1. All powers of a present in d being present
in kd, with no further terms coinciding, d A kd.
If x = 2, set d = ELo a6*. Then kd = EL0 aM+1 + ELi W + Ej=26i+2-
Now a4l+2 = ct, so that a 2a appears. But 2a = a2l+l, and 2a2l+1 = 1, so that
d < kd in this case.
Finally, if x = 1, h = 4i and r = 2i. This case cannot occur, since (h, r) > 1.
Thus, (k) is not a hyperoval in PG (2, 2h).
All that remains now is r = i + l + 2 with l = i 1.
Proposition 4.13 Let k = a + a1 + a2l+x, where a = 2l with al+l+2 = 2.
Further, let h = 3i + l + x, where l = i 1 and 1 < x < l. Then (k) is not a
hyperoval in PG (2, 2ft).
Proof: We consider five cases on the values of x. If x > 5, let d = Y,t:oV +
Et+:
b
i+x1
i+x4
a
+ a2l+2x~6. Then kd = a* + Â£12 a" + a2-5 + ESSE +
2x5 Q,2i+2x4
a2t+2x-5 _|_ q,3j+2x-6 _|_ Efe^3i+2x-4 It Is possible that a 2a
term appears, in which case 2a2l+2x~4 = ax~2, which cannot otherwise appear.
Necessarily, however, a 2a2l+2x~b = ax~z appears, and 2ax_3 = a2l+x~2 and
2a2l+x~2 = 1 so that d ^ kd.
If now x = 4 consider d = Eb=ofti" + a2l+1- Then kd = EfLo0^1 +
ELi W+Et2 abi+4+a2i+2+a3i+1+a4i+5. Now a4l+4 = a, so that a 2a = a2l+2
appears. Now 2a2l+2 = 1, so that d kd in this case.
When x = 3, h = M + 2 and r = 2i + 1, so that (h,r) > 1, and this case
does not occur.
64
When x = 2, consider di-i. Now 2a2l+2 + 2a2l+3 appears in kdi_x; however,
2a2l+2 _|_ 2a2l+3 + a2 = a2l+3 + a3, neither of which terms otherwise appear, so
that di^i -< kdi-i.
Finally, when x = 1, let d ab + a2j_1. Then kd = JJb=o ab + 2a +
Ylll=i+i ab + 2a3*-1 -f a3'. Now 2a3-1 + 2al = a1 + aZl+l so that d < kd.
Thus, S (k) is not a hyperoval in PG (2, 2h).
Lemma 4.14 Let k = a+o:l + Q!2l+x, where a = 2l with (io, h) = 1 and ar = 2.
Further, let h = 3i + l + x, where l < i and 1 < x < l. Then (k) is not a
hyperoval in PG (2, 2h).
Proof: Table 4.1 illustrates how Propositions 4.1-4.13 together classify this
case.
With all possibilities covered, the result follows.
4.2 Case 2
In our examination of case 2, we are very quickly able to use values of d
similar to those used above to severely restrict the parameters m, n, and j. Our
first proposition states that m and n may not simultaneously be greater than
one.
Proposition 4.15 Let k = a + a2 + aj, where ar 2. Further, let h =
mj + 2n + 1, where 2n + 1 < j. If m > 2 and n > 2, then (k) is not a
hyperoval in PG (2, 2h).
Proof: Define the family {datb} of values for d in the following way. If a < n b,
hand, a > n-b, da, =
65
Table 4.1: Case 1 Classification Outline Conditions Proposition
r < min {x, l x} 4.2
min {x, l x} + 1 < r < i + x 2 4.1
r i + x 1 4.9
i+x
t = i + Z-t-1, 2 4.10
t = i 1 1, l % 2 4.11
r i + l + 1, l = i 1 4.12
r = i + l + 2, l < i 1 4.1
r = i + l + 2, l = i 1 4.13
i + l + 3
r = 2i + l 1, l < i 1 4.1
r = 2i + l 1, l = i 1 4.6
2i-\rl
r 2i + l + x + l 4.4
t = 2i 1 x 2, l i 1 4.1
r = 2i + l + x + 2, l = i 1 4.5
2i + l + x + 3
3 i + l < r 4.2
Note that dn_b,b has been defined in two different ways. When we wish to use
the value defined under the a < n b assumption, we will denote it by d~_bb,
and when we wish to use the value defined under the a > n b assumption, we
66
will denote it by d^_b b to avoid confusion.
We now determine the value kd^b for each of the definitions of daIf
a < n b, we find the following value for kd^b.
2a+l
2(n6)+3
nb 1
kda,b = Y olc + 2a2a+2 + Y aC + J2 ^ + J2 ^+2C+1
c=0
j+2(n+l)+l
c=2a+3
n1
c=0
+ E c+ E
OL
2j+2c+\
c=j+2(nb)+l c=nb
m1
+ Y (acj+2n+1 + acj+2n+2 + acj+2n+3) .
c= 2
The exponents which appear are distinct whenever 1 < b < n 1. Every power
of a present in dab is present in kdatb, and the only duplicate term which appears
is 2a2a+2.
In a similar fashion, we compute the value kda,b fr the case in which a >
n b. In this case, we obtain the following:
2(n6+1) n-6-1 j+2a+l j+2a+3
kda^ a+ aj+2c _j_ E ' + 2++2+2+ Â£
c=0 c=0 c=j+2(n-6) c=j+2a+3
a
n1
m1
+ E
a
2j+2c
E
a
2j+2c+J
+ E(
acj+2n+l _j_ acj+2n+2 ^ acj+2n+3^ _
c=nb ca c= 2
Once again, all of the exponents are distinct (so long as 1 < b < n 1), all
powers of a that appear in da,b also appear in kdaib, and the only duplicate term
which appears is 2oP+2a+2.
We can now determine the values for r that are not eliminated by each of
do,6) dn-b,b' dn-b,bi anCl dn,6-
67
For d0)b the values that remain are
1 < r <2(n 6) 1
r = J 2
r = j + 2c 1 : 0 < c < n 6 1
j + 2 (n 6) 1 < r < j + 2n + 1
r = 2j + 2c 1 : n b < c < n 1
r = cj + 2n 1 : 2 < c < m 1
r = cj + 2n: 2 < c < m 1
r = cj + 2n + l : 2 < c < m 1
mj + 2n 1 < r < mj + 2n.
For d~_b b the values that remain are
r = 1
r = j 2c : 1 < c < n 6+1
j 1 < r < j -f 2b + 1
r = 2 j + 2c 1 : 0 < c < b 1
r = cj + 2b 1 : 2 < c < m 1
r = cj + 2b : 2 < c < m 1
r = cj + 26 + 1 : 2 < c < m 1
mj + 26 1 < r < mj + 2n.
68
For b the values that remain are
1 < r < 2b + 1
r = j ~ 2
r = j + 2c 1 : 0 < c < b 1
r = cj + 261: 1 < c < m 2
r = cj + 2b : 1 < c < m 2
r = cj + 26 + 1 : 1 < c < m 2
(m 1) j + 2b 1 < r < (m 1) j + 2n + 1
r = mj + 2c1: b < c < n 1
77V? + 2n 1 < r < mj + 2n
For dn b the values that remain are
r = 1
r = j 2c : 1 < c < b + 1
r = cj 1 : 1 < c < m 2
r = cj : 1 < c < m 2
r = cj + 1 : 1 < c < m 2
(m 1) j 1 < r < (m 1) j + 2 (n b) + 1
r = mj + 2c 1 : 0
mj + 2(n b) l
Under the assumption that 1 < b < n1, examining the remaining values for
these values of da^ reveals that only the following values for r are not eliminated
69
by at least one of the da^.
r = 1
r = j ~ 2
r = j- 1
r = j + 1
r = 2 j + 1
r (m 1) j + 26+ 1
r = mj + 2n 1
r = mj + 2 n.
We eliminate these values for r in turn. For r = 1, the only duplicate term in
kdnis 2oJ+2,l+2 = cF+2n+3, and 2aJ+2"+3 = ct7+2n+4, which does not otherwise
appear when j ^ 2n + 2. When j = 2n + 2, 2cF+2n+4 = aj+2n+5, which does not
appear, so that dn*
If r = j 2 and j 7^ 2n + 3, the only duplicate term in kd~0 is 2a2n+2*
or 3a2n+2, and 2a2n+2 = a2-74-2"-2, which does not otherwise appear, so that
d~0 ^ kd~ o- When j = 2n + 3, let d = X^c=o a2 + Qfc-7+2n+1. Then kd =
Â£c=o V+2c*2n+3+a2"+4 + Â£c=i cF+2c+2a7+2"+2+7+2"+3 + Â£^=21 (aC7+2n+1 +
Q,cj+2ri+2 _|_ acj+2n+3^ Now 22ri+3 = Qf7"1"2"4"1, so all terms of d appear in kd, and
2od+2n+2 = a2-74"2", which does not otherwise appear so that d ^ kd.
Now if r = j 1 and j > 2n + 3, the only duplicate term in kd0^0 is
2a2 = a-74-1 term. Now 2a-7+1 = a2-7, which cannot otherwise appear in kdo^,
so that do,o ^ kdo Q. If j = 2n + 3, let d = Â£^o a2c + 1l!c=2 ac^+2n+l. Then
kd = Â£c2:t2 c + 2a2n+3 + a2n+A + Â£"Â£' cF+2c + (aC7+2n+2 + ac-*+2n+3) +
Â£Â£3 a0-74"24-1. As 2a2n+3 = o;t+2n+2 anc[ 2o;t+2n+2 Q,2j+2n+i^ ^ ^ Finally,
70
if j 2n+2, let d = EEo <*2c+E / otCJ+2n+l, so that kd = Ec^o1 c+2a2n+2+
EEi aj+2c + aj+2n+2 + ctj+2n+3 + J2Z~2 (uCJ+2n+1 + acj+2n+2 + acj+2n+3). Since
2a2n+2 = ot?+2n+1, d X kd.
If r = j + 1, consider the value d0,i- Since 2a2 = od+3 and 2aJ+3 = a2j+4,
which cannot otherwise appear, d0,i ^ A;d0,i.
When r = 2j + 1, take d = Ec=o ^ + a2"-1 + a-7+2n_1 + EEi^ aCJ+2n+1, so
that fcd = EcE* c + 2a2" + E=o qJ+2c + aJ+2"_1 + aJ+2" + aj+2n+1 + 0i2i+2n-1 +
E^1 (cj+2n+2 + o;ct+2n+3) + Er=3^+2n+1. Since 2a2n = a2i+2n+l, d r< fcd.
For r = (m 1) j + 2n 1, let d = 1 + a + Q-(m_1W+2Tl+1. Then kd =
l+a+2a2+a3+aJ+aJ+1+a(m~lb'+2"+2Wm_1)j+2"+3. But 2a2 = c*(-W+2n+i
so that d ^ fcd.
If r = mj + 2n 1, take d 1 + a. Then kd = a + 2a2 + a3 + aJ + aJ+1,
and 2a2 = 1, so that again d ^ fcd.
Finally, when r = mj + 2n, if j > 2n + 4 set d = E"=o q2c + EEi* &cj+2n+1-
So kd = E2o3c + Ec=oj+2c + 2a-7+2n+2 + a7+2"+3 + EE21 (acj+2n+1 +
acj+2n+2 _|_ 0,^4-271+3^ now 2ai+2n+2 a-?+2"+1, so that d < kd. On the other
hand, if j < 2n + 4, set d = 1 + a2n+2 + YZ?=i acj+2n+1 so that kd = 1 + a + a2 +
a2n+3 + a2n+4 + aj + 2aj+2n+2 + aj+2n+3 + Y^ (aci+2n+l + acj+2n+2 + ^+277+3)
Now a-7 G {a2n+2, a2n+3, a2n+4}. As 2a2"+4 = a2n+3 and 2a2n+3 = a2"+2, and
0,271+2 (p)es no^ appear apart from a?, a2n+2 appears in each case. Further,
2a.i+2n+2 = ctJ+2n+1, so that d A kd.
Thus, for every choice of r, some value of d was found for which d < kd, so
that, if m > 2 and n> 2, @ (k) is not a hyperoval in PG (2, 2h).
We next consider the situation in which m = 1 and j > 2n + 3.
71
Proposition 4.16 Let k = a + a2 + a3, where a = 2l, ar = 2, j > 2n + 3, and
h = j + 2n + 1. Then if @ (k) is a hyperoval in PG (2,2h~), it is a translation
hyperoval.
Proof: As defined in the proof of Proposition 4.15, consider da0. Note now
that kdafi can be simplified to a + 2a2a+2 + a + Sc=o aj+2 +
aJ+2c+1. Notice also that only the following values for r are not ruled out
for a particular a:
l
r j 2c : 1 < c < a + 1
r = j + 2c 1 : 0
j + 2 (n a) 1 < c < j + 2n.
For o = 0, this equates to the following values for r:
1 < r < 2n + 1
r = j -2
r = j + 2c 1 : 0 < c < n 1
j + 2n 1 < r < j + 2n
while for a = n, this leaves the following values for r:
r = 1
r = j 2c : l
j 1 < r < j + 2n.
72
Combining these two lists, only the following values for r remain (noting that
the lists are identical if n = 0):
r = 1
r = j 2c : 2 [?r2~1j < c
r = j 2
r = j + 2c 1 : 0 < c < n 1
j + 2n 1 < r < j + 2n.
We again rule out each of the remaining cases in turn. When r = 1, consider d~0.
Since 2a2n+2 a2n+3 and 2a2n+:i a2n+4 (which may not otherwise appear in
the product) and 2a2n+4 = 2n+5, where a2n+5 does not appear in the product
kd~o otherwise, d~0 < kdnt0. This fails only if n = 0 and j 4, in which case,
k = 2 + 22 + 24, and 1 j^ = 22, so that 2> (k) is a translation hyperoval in
PG (2,4).
For r = j 2c, we consider two cases depending on the parity of j. If j
is even, consider d0i0. In the product kd0t0 a 2a2 appears. Now with j even,
r is also even. But every power of a greater than 2 in d is odd. Furthermore,
h is odd, and the powers a^+2n~2c for 0 < c < n 1 are all absent from the
product kdofi- Since this r value is necessarily less than 2n + 1, if new powers
of a continue to arise from the 2ct2 through continued carrying, one of these
powers must be of the form a^+2n~2c with 0
This again fails for n = 0, however, n = 0 implies that j 1 < 2, so that j = 3,
h = 4, and | = 2, so that Q> (k) is a hyperconic.
If j is odd, and j 2c < 2n + 1, we use d = 2 a29 + o?^n+c~Lr )+1 +
a2n+l. Then kd = a9 + 2a2n+2c~j+1 + a^n+2c-j+2 + a2n+2 + a2n+3 +
73
X!g=o 2 +29 + cr'+2(n+c~i2_)+1, with all powers distinct. But 2a2n+2c~i+1 =
an+l, so that d ^ kd.
On the other hand, if j 2c 2n + 1, we divide into two further
cases. If j > 4n + 4, kdoto has a 2a2 term. Now 2a2 = a2n+3 and
2a2n+3
= a4n+4. Since 4n + 4 < j, this does not otherwise appear, so
that d0,0 Z. kd0io- If j < 6n + 3, consider d = ^1 + a^ X]g=o 1 0Z9
Here, kd = ^1 + a^ 2"a;9 + 1 aj~%+29^J. Notice that with
j < 6n + 3, all of the exponents in the second factor are less than |. Now j |
is equal to either 2 [|J 2n 1 or 2 [|J 2n, and in either case, 2a= a%,
so that all terms of d arise in kd.
Now, if r = j 2 we consider four cases. For j > 2n + 6, we consider
d^0 In the product kdn (h 2a2n+2 = aJ+2n, and 2a3+2n = a-7-3 which does not
appear, so that d+0 < kd0. If j = 2n + 6, let d = 1 == J2c=oa2c+3 so that
kd = 1 + a + 2a2 + ^=4o:c + YZUZo ad+2c+3. Now 2a2 = cd and 2aJ = a3,
so that d < kd. This fails when n = 0, in which case ^ = 6 and @ (k) is
a Segre hyperoval. When j = 2n + 5, h = 4n + 6 and r = 2n + 3 = |, so
that (r, h) ^ 1, a contradiction. Finally, if j 2n + 4 let d = a2c- Then
kd = 2a + JZiZ3 qC + 2ad + X]c=i a2c+2- But 2a? = a, 3a = a + a-7"-1, and
2aj~1 1, so that d ^ kd.
When r j + 2c 1 with 1 < c < n 1, we use d = Z2gZo ft2c + a2(n~c^+l,
so that kd = ZZfZZ^1 a9 + 2a2(n~c'l+2 + a2^n_^+3 + ZZI^Zq o?+2a + ad+2(n_c)+i_
Now 2a2(n~c)+2 = 1, so that d ^ /cd.
If r = j 1, we consider several cases. If j > 2n + 6, consider d_i;o (note
that with 0
74
a 2a2n appears. Now 2a2n cP+2n~1 and 2aj+2n~l = aj~3. Since j > 2n + 6,
j 3 > 2n + 3, so that d_1;0 A ^dn-i,o- Now, if j = 2n + 6 and n > 1, set
d = X^c=o a2c + a2_3 + o:2"+1, so that kd = X^c=o3 c + 2a2"-2 + a2n-1 + a2n+2 +
a2"1-3 + ^02 a3+2c + a-7-1"2-3. Now 2a2n_2 = aj+2n~3 and 2cr7+2n~3 = a2n+1,
so that d < kd. If j 2n + 6 and n = 1, set d = 1 + a2 + a6, so that kd =
a + Q;2 + 2a!3 + a:4 + a:7 + 2ai8 + a10. But 2a3 = a10 and 2a10 = a6, while 2a8 = a4
and 2a4 = 1, so that d ^ kd. If j 2n+5, then h = 4n+6 and r = 2n+4, so that
(r, h) ^ 1, a contradiction. Finally, if j = 2n + 4, consider d = 1 + ]>Zc=i a2c+1,
so that kd = 1 + a + 2a2 + c + 22n+4 + a2n+5 + aj+2c+1. Now
2ct2"+4 = a2, 3a2 = ct2+5 + a2, and 2a2+5 = a3, so that d -< kd.
We next consider r = j + 2n l. As in the proof of Proposition 4.15, we use
d 1 + a, so that kd = a + 2a2 + a3 + a? + cP+1. So long as n ^ 0, Of-7+1 ^ 1,
in which case 2a2 = 1 and d < kd. If n = 0, consider d = 1 + a2 + a3. Then
kd = 2a + 2ct2 + a3 + 2a4 + a5 + a?. Now 2a4 = a2, 3a2 = a2 + 1] 2a = cr7, and
2cP = a-7-2, which no longer otherwise appears unless j G {3, 4, 5, 7}. If j = 3,
1 k = 2 and @ (k) is a hyperconic; if j = 4, 1 k = 6 and @ (k) is a Segre
hyperoval; if j = 5 or j = 7, (r, h) 2, a contradiction.
Finally, if r = j + 2n, the values of d used in the proof of Proposition 4.15
work equally well with m = 1 for all values of n.
Thus, in all cases, either a value of d was presented with d ^ kd, or @ (k)
was shown to be a translation hyperoval or a Segre hyperoval.
We next deal with the situation in which j = 2n + 3. In such a situation,
we have the following result.
Proposition 4.17 Let k = a + a2 + a^, where a 2*, ar = 2, j = 2n + 3, and
75
h = j + 2n + 1. Then if (k) is a hyperoval in PG (2, 2h) it is a hyperconic.
Proof: We split into cases for different values for r. Note that h = A(n + 1),
so we need only consider odd values for r. If r = 1, then consider d~0. A
2a2n+2 term appears, creating a 3a2n+3 term, which in turn creates an ot2n+i
term, which does not otherwise appear unless n 0, in which case j = 3, and
1 (a + a2 + a3) = a. But a = 2, so that @ (a + a2 + a3) is a hyperconic in
PG (2,24). Thus, for n > 0, d~0 ^ kd~0.
Now, if 3 < r < 2n 1, consider d = 1 + a + V'T! .r+i a2c+1. Then kd =
C~ 2
l+a + 2o2+a3 + Ec=r+3c + 2^2rl+3 + 2n+4 + EcJ4i J+2c+1. But 2a2 = ar+2,
while 2o2+3 = a2n+r+3, which does not otherwise appear, so d < kd.
When r = 2n+l, we set d = 1+a, so that kd = a + 2Q;2 + Q!3 + a2Tl+3 + Q;2n+4,
and 2a2 = o2"+3, giving rise to a 2o2+3 = 1, so that d z< kd.
On the other hand, when r = 2n + 3, we use d0i0. Then 2a2n+3 = a2, and
3a2 = a2 + a2n+5, which does not otherwise appear unless n = 0, in which case,
k = a + a2 + a3 and h = 4, so that k = 2 + 22 + 23, and 1 k = 2, making @ (k)
a hyperconic in PG (2,24). For n > 0, though, d0,o ^ kd0p-
If 2n + 5 < r < An + 1, define d 1 + V r-2n-i o?c + ct2n+1, so that
2
kd l+a + a2 + E2^2nc + 2a2n+2 + 2a2n+3 + E"=^i j+2c- But 22"+2 =
olt~2"-2, while 2a2"+3 = ar~2n~x, neither of which terms otherwise appear, so
that d -< kd.
Finally, if r = 4n + 3 = j + 2n, the same value of d as used in the proofs of
Propositions 4.15 and 4.16 still works.
Thus, in all cases in which (k) was not a hyperconic, a value of d with
d -< kd was exhibited.
We obtain a similar result when j = 2n + 2.
Proposition 4.18 Let k = a + a^ + cP, where a = 2l, ar = 2, j 2n + 2, and
h = j + 2n + 1. Then if 2 (k) is a hyperoval in PG (2, 2h), it is either a Segre
hyperoval or a Glynn hyperoval.
Proof: As usual, we divide into cases based on the different possible values for
r. Notice that n > 1, as n = 0 implies that j = 2, a contradiction.
For r = 1, the value d~0 as defined in the proof of Proposition 4.15 gives
a product with the only repeated term being a 3a2n+2. But 3o2n+2 = o-2n+2 +
a2"1'3, 2ai2ri+3 = a2"+4, and 2a2n+4 = a2n+5 which does not otherwise appear so
long as n > 2. On the other hand, when n = 1, we obtain k = 2 + 22 + 24. But
then 1 Â£ = 6, so that 2 (k) is equivalent to the Segre hyperoval in PG (2,27).
If 2 < r < 2n 1, define d = 1 + a + ^"=j- 2 ^ ar+2c + a2n+1. Then kd
l+a + 2a2 + a3 + ac+{2+tf [=JJ) a2n+2 + 22"+3 + aJ+2c+r.
Now, since [2 [|JJ is zero or one as r is odd or even, the coefficient on a2n+2 is
either 2 or 3. But then a 2n+2+r term arises, which does not otherwise arise, and
the 2a2n+3 likewise gives rise to an a2n+3+r term which also does not otherwise
arise. Finally, 2a2 = ar+2 so that d < kd for these values for r.
Now if r = 2n, we examine d~0, and notice that a 3a2n+2 term arises. But
3a2n+2 = a2n+2 + ai4n+2, 2a4n+2 = a2n_1, and 2a2n_1 = a4"-1, which does
not otherwise arise, so long as n > 3. Now, if n = 1, k = 24 + 2 + 22 since
h = 7, so that 2 (k) was earlier shown to be equivalent to the Segre hyperoval
in PG (2,27). If, on the other hand, n = 2, k = 23 + 26 + 27 with h = 11. Then
1 | = 23 + 26, where (23)4 = (26)2 = 2, so that 2 (k) is equivalent to the
77
Glynn hyperoval (a + 7) in PG (2,211).
For r = 2n + 1, d = E"=o q2c giyes a product of kd = EcHt* a<: + 2a2n+2,
and 2a2"+2 = 1, so that d ^ fcd.
When r = 2n + 2, consider d = E"=o a2c + ce2n_3 + a2"-1 + a2", so that
kd = ESc/V + 2a2"-2 + a2"1 + a2n + 2a2"+1 + a2n+2 + Ec=o <*2n+2c+2 +
0,471-1 _|_ a4n+1 4- ain+2. Now 2a2"-2 = a4", which does not appear elsewhere in
the product. Furthermore, 2a2"+1 = 1, so that d < kd, as desired. This fails
precisely when n < 3. But when n = 1, we again have k = 2 + 22 + 24 and
h = 7, giving a Segre hyperoval, and when n = 2, h = 11, and k = 2 + 22 + 24,
where ^ = 22 + 26 + 27, or 3a + 4, a Glynn hyperoval.
If r = 2n + 2c + 1 with 1 < c < n, set d = YZ^Zo0-29 + a2^"-c)+1. Then
kd = E2LV)+1 a + 2a2("-c)+2 + a2("-c)+3 + ZgZo a2n+2+2 + a4"-2c+3. But
2a2(n-c)+2 = so that d ^ kd.
If r = 2n + 2c + 2 with 1 < c < n, set d = 1 + J2g=c 0(29+1 > so that
kd = 1 + a + a2 + Eg=2c+2 a9 + 2a2"+2 + a2n+3 + YZgZl a2"+2ff+3. But since
2a2"+2 = a2c+1, d < kd.
Having covered all possibilities for r, it follows that in this case any hyperoval
is a Glynn hyperoval or a Segre hyperoval.
We finally address the case in which m >2 and n < 2 in the two propositions
that follow.
Proposition 4.19 Let k = a + a2 + a-7, where a = 2l, ar = 2, and h = mj +1.
Then if @ (k) is a hyperoval in PG (2, 2h), it is a translation hyperoval or a
Glynn hyperoval.
78
Proof: We first consider which values are ruled out by g^0, noting that kd,Q 0 =
l + a + 2a2 + a3 + a-7 + (a^+1 + + oc-7+3)- Subtracting all powers
of a in the product, we are left with r G {1, j 2, cj 1, cj, cj + 1, mj 1, mj}
where 1 < c < m 1. But further for j > 4, if r = 1, 2a2 = a3, and 2a3 = a4
which does not otherwise appear.
Now, if j > 6, and r = j 2, notice that 2a2 = aj and 2(P = a2j~2, which
does not otherwise appear in the product.
If j > 5, and r = cj + 1, for 1 < c < m 1, 2ct2 = a
o2c-?+4. If 2cj + 4 < h, this term does not otherwise appear, while if 2cj + 4 > h,
2cj + 4 gj + 3 for some value of g. Now there exists some first multiple of
r (say q) such that 2 + q ^ 3 mod j. But then, necessarily, 2 + r (q 1) = 3
mod j, so that 2 + qr = 4 mod j. Furthermore, a2+9m appears for all m < q,
but does not then appear for m = q. Thus, d -< kd.
This analysis of 0 fails for j = 3, when an additional doubled term appears;
however, the values remaining comprise all values between 1 and h 1 when
j = 3, so that the following analysis is in fact complete.
We now analyze the remaining values for r.
If r = 1 and j = 3, let d = J2T=o ct3c+a3^m^+1. Then kd = ac+
2a3(m-2)+2+2a3(m-l)+2a3(m-l) + l+a3(m-l)+2+a3m_ Now 2a3(m-2)+2+2a3(m-1) +
2a3(m-1)+1 + a3m = a3('m~1'> + a3(m-1)+i + 1, so that d ^ kd.
If, on the other hand, j = 4, set d = 1 + a2 + o;4c+1, so that kd
^3=0 ac + 2a4 + 2ct6 + a7 + Y^=2 (a4c + 4c+1 + ct4c+2). Now 2a4 = a5, while
2a6 = a7 and 2a7 = a8, which does not otherwise occur in the product so that
d -< kd.
79
Now if r = j 2 and 4 < j < 5 (since 3 2 = 1 we need not consider
j = 3 again), consider d = E^2^ + c^m-2^+1 + a^m_2^+3, in which case
kd = ZZo2 (aCJ+1 + c^'+2) + ZZl1 + u{m~2)j+2 + a^-2^ + c^-W+i +
a(m-2);+4 _|_ Q,(m-2)j+5 _j_ a(m-l)j+3_
If j 4, the multiple powers that appear are precisely 2a^rn~2^+2 = ,
2cdm-2^+4 2with 3a^m_1^ = + Q/i771-1)^2 which does not
otherwise appear, and 2a^m~1^+1 = a^m_1h+35 which already appears, so that
2a{m-\)j+z j an(j d < kd.
If j = 5, the multiple powers that appear are precisely 2cdm-2h+2 = a(m~i)j
and 2a^m~1^, so that the resulting 3a^m_1^ = + a(m_1h+3. But
2Q,(m-i)j+3 so that d -< kd.
If now r = j 1 and j = 3, we use d = ZT=o ^C + a3m_4, so that
kd = Ec3=r4c+23m-3+3m-2+3"1-1. But 2a3m~3 = a31 and 2a3771-1 1,
so that d < kd.
If j = 4, we note that if m = 2, h = 9 and j 1 = 3, so that this case
cannot occur as (3,9) ^ 1. We use d = EEo3q;4c + a4^m_3^+2 + Q,4(m_2)+2, so
that fed = Â£r=o3 (a4c+1 + a4c+2) + E? 4c + a4^-3)+3 + a4(m:2) + a4^+2 +
Q,4(m-2)+3 _|_ _|_ 0,4777-2 now a 2a4(77i-2) ^4(771-2)4-3 term appears, but
2Q;4(77l-2)+3 -- a4m_2; and 2o;4m2 = 1, so that d X A;d.
Finally, if j > 4, consider d E^o* ^ + a(m-2^+2, in which case kd =
Z7=o (<*cj+1 + acj+2) + Ei <*cj + (m_2)i+3 + aim~2)j+4 + a^m^j+2. Here, a
2a{m-i)j+2 appears, but this simplifies to 1, so that d -< kd.
When r = cj 1 for 2 < c < m 1 and j = 3, we set d =
Er=oC_1 39 + a3(m"c)-1 + Q!3(m_c)+1. Then kd = EsLTcM a + 2a3(m~c) +
80
a3(m-c) + l + 2a3(m-c)+2 + a3(m~c)+3 + a3(m-c)+4_ Nqw 2aHm-c) = a3m-l which
does not otherwise appear, and 2a3(-m~c'l+2 = 1, so that d < kd.
On the other hand, if j ^ 3, set d = Y^=o 0/99 + a;(m~ch'+1) so that kd =
EJTcT1 (9j+1 + a"+2) + Er=T+1 a9j + a(m_cb+1 + 2a(m-cW+2 + cdm-c)J+3 +
o;(m_c+ib'+i_ gince 2cdm-ch'+2 = 1, d ^ fcd.
If r = j and j ^ 3, d = 1 + a2 + EEV 0/09+1 gives the product kd =
1 + a + a2 + a3 + a4 + a? -f 2aJ+2 + aJ+3 + EE^ (Q;C:+1 + + aC9+3). Now
2aJ+2 + E^1 olC9+2 = a, and 2a = aJ+1 so that d -< kd. (If j 4, 2a4 = a2j,
which does not otherwise appear.) When j = 3, k = 2 + 2m+1 + 22m+1. Notice
then that | = Ec=m 2C, so that 1 Â£ = 2m, where (m, 3m + 1) = 1, and ^ (fc)
is then a translation hyperoval.
Now if r = cj with 2 < c < m 1, for j 3, consider d = 0/99 +
a(m.ci)j+i, where kd = Y^=o~2 (aS'7+1 + a9^+2) + "Y^=\+l 0/99 + cdm~c~:Lh+1 +
2a(m_c_1h+2 + Q!(1-c-1)f+3 _p 2a^m_c^+1 + a(rn_ch'+2_ Now 2a^m_c_1^+2
Q,(mi)j+2 ^jch (joes not otherwise appear in the product, and 2a^m_c_1^+1 = 1,
so that d -< kd.
When j = 3 and c > 2, consider d = Y^=o~2 0/39 + a3(m~c_2)+1 +
a3(m~c~i)+i-f a3(m-c), so that kd = E^Ti~C~2^+1 ct9 + 2a3(m-c_2)+2 + 2a3(m~c~1) +
a3(m-c-l) + l _|_ a3(m-c-l)+2 _|_ a3(m-c) 2a3(m_c)+1 _p Q,3(m-c)+2 a3(m-c+l)_ Now
2a3(m-c-2)+2 = a3(m-2)+2^ 2a3(m~c~1') = a3^m-1\ and 2a3^m_c^+1 = 1, so that
d ^ kd.
Finally, if j = 3, r = 6, and m > 6, consider d = ESu)6 a3c + a3^m_6^+2 +
a3(m_5)+1 + o;3(m_4) + Q,3(m3)+2^ jn this case; kd = X)^_6')+2 ac + 2a3^m_5^ +
a3(m-5)+l _p 2o;3(m-5)+2 -p Q,3(m-4) -p 2ct3(m_4) + 1 + a3(m-4)+2 _|_ ^3(771-3) a3(m-2) _|_
81
a3(m~2)+i + a3(m-2)+2 j\jow 2c*3h"~5) = o;3(m-'3) and 2c*3h"-3) c*^"1-1) which
does not otherwise appear. Further, 2c*3h"-5)+2 = ct3^-3^2, which does not
otherwise appear in kd but does appear in d. Finally, 2c*3hn-4)+1 = a3(m-2)+1,
and 2a3(m~2)+1 = 1, so that d ^ kd.
In the case that r = cj, this leaves only the possibility that j = 3, r = 6,
and m < 6. Now if m is odd, h is even and (r, h) > 1, so that we need only
consider m = 4 and m = 2. If m = 4, notice that k 27 + 29 + 2U, with h = 13.
Then | = ]Cc=2 2C + ]Cc=9 2C, so that 1 | = 22 + 27 + 28 = 3cr + 4, making
@ (k) a Glynn hyperoval. When m = 2, notice that k = 24 + 25 + 26 and h = 7,
so that 1 k = 24 and ^ (fc) is a translation hyperoval.
We next consider r = cj + 1 when j < 4. Here let d YgYtf1 ft9J +
a(m-c)j-2j so kd = ^rgn~o~l (9j+1 + a9j+2)+Er=7~X a^+(m-c)^1+2a(m-cb +
^(m-c+ip-2 gut then cj j 2 6 {(m c 1) j + 1, (m c 1) j + 2},
2c*h"-cb-1 = c*^ which does not otherwise appear, and 2cdm_ch = 1, so that
d -< kd.
When r = mj 1, as in the proof of Proposition 4.15, d = 1 + a suffices
so long as j 3. If j = 3 and m > 2, let d = 1 + a3 + c*5, so that kd
Ec=i Q^ + Sa;6 + c*7 + c*8. Now 2c*6 = c*4, 2c*4 = a2, and 2c*2 = 1, which does not
otherwise appear, so that d z< kd. If, on the other hand, m = 2, A: = 22 + 23 + 26,
and 1 k = 22 + 24 + 25 = 3cr + 4, so that @ (k) is a Glynn hyperoval.
Finally, when r mj and j > 4, choose d = 1 + aj~l + YZ2 aC9+\ so that
kd = 1 + c* + c*2 + 2c*J + c*2j_1 + Y=2 (&cj+1 + c*c:,+2 + c*CJ+3). Now 2c*J = cP-1,
so that d z< kd. When j = 4, let d = EEo1 ^C + c*4m_5 + c*4m_3, so that
kd = YZo1 (4c + c*4c+1 + c*4c+2) + c*4-4 + c*4m~3 + c*4m~2 + 2c*4"1'1 + c*4m.
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Now ^Zc=4m-4 ac + 2a4m~1 = o4m_5, which does not otherwise appear, and thus
d kd. On the other hand, whenj = 3, k = 2h~z+2h~2-\-2h~l, and 1 k = 2h~z.
Since 3 does not divide h, (k) is a translation hyperoval.
Thus, having presented either a d with d kd, or an equivalence to a known
hyperoval in all cases, the result follows.
We finally classify the hyperovals with n = 1 and m >2.
Proposition 4.20 Let k = a + a2 + od, where a = 2l, ar = 2, j > 3, and
h = mj + 3. Then if (k) is a hyperoval in PG (2,2h), it is a Glynn hyperoval.
Proof: We first examine the four relevant da b values with the same analysis
as in the proof of Proposition 4.15. If j > 5, do,o leaves the following values for
r:
1 < r < 3
j-2
cj + l
mj + 1 < r < mj + 2
If j 7^ 5, df o leaves
r = 1
r = j 4
r = j -2
cj l
raj 1 < r < mj + 2
83
If j > 4, d~t0 leaves
r 1
r = j 2
cj l
mj j + 2 < r < mj j + 3
r = raj 1
mj + 1 < r < mj + 2
Finally, if j >4, leaves
r = 1
j 2 < r < j
cj + 1
2j l = r
mj + 1 < r < mj + 2
84
Putting this together, at most the following values remain to be classified:
r = 1
j 2 < r < j 1
r = j = 5
r = 2j 1 = 9
r = cj 1 2 < c < m 1 j = 4
r cj 1 < c < m 1 J = 4
r = cj + l 1 < c < ra + 1
mj 3 < r < 2 j = 5
raj 1 < r < mj j 4
r mj + 1
r = mj + 2
We classify these in turn.
If r = 1, consider d0,o- In this product, we have 2a:2 + a3 + aA + a5 = a6,
which does not appear unless j = 5 or j = 6. If j = 5, 2a6 = a7, which does
not appear, and if j = 6, 2a6 + a7 = a8, which does not appear, so d z< kd.
For j / 5 and r j 2, consider d^0. We obtain a 2a4 = a?+2 (or
3a4 = a4 + ai+2 if j = 4), and 2a4+2 = a2-7 which does not otherwise appear
unless j = 4. Now if j = 4, 2a2-7 = a2-74"2 which does not otherwise appear so
that d ^ kd.
When j = 5, consider d = 5c + a5(m_P+1, in which case the product
kd = Â£r=o2 (<*5c+1 + a5c+2) + Er=" q5c + a5(m_1)+1 + 2a5(-1)+2 + ^(m-D+3 +
a5m + a5m+1. Now 2a5(m_1)+2 = a5m and 2a5m = 1, so that d < kd. For
r j 1 and j > 4, consider d = a>+ a^m_2^+2 + a^m_2^+4 + a^m_1b+2,
85
so that kd = ES2 (acS+I + <*+2) + EE',' o'3 + ESm'+s c + a<"">W+2 +
a(m-i)j+3 _|_ 2a(m-l)j+4 + aT?y'+2. Now 2a^m_1^+4 = 1, and, if a 2ahn~1b term
appears, 2cdm_1) = a-7-1, which cannot otherwise appear with j > 4, so that
d -< kd as desired.
On the other hand, if j 4. consider d = Y^T=o a4c + a4m_2, so that
fcd = E^o1 (4c+1 + 4c+2) + E^Ti1 a4c + a4m"1 + 2a4m + a4m+2. Now 2a4m = 1,
and d ^ kd.
Now when r j and j = 5, consider the value d0,o- In the product fcdo.o, we
obtain a 2a5 term and a 2ct2 term. Now 2a5 + E^=2 ^c = (where the sum is
assured to appear in the product), and now 3a2 a2 + a7, where a7 does not
otherwise appear, so that do,o ^ kd00.
If now r 2j 1 and j = 5, let d = Ecl^2 a5c + a5^m_2^+2 + a5^m_2^+3. Then
fed = Er02 (5c+1 + 5c+2) + EEf ft5c + a5(m-2)+3 + 2a5(m"2)+4 + 2a5(m_1) +
Q,5(m-i)+2_j_Q,5(m-i)+3 ]\jow 2a5(m-2)+4 j. 20,5(771-1) = a, 2a = a10 which does
not otherwise appear if m = 2, 2a10 = a19, which does not otherwise appear.
Thus, d < kd.
For the case in which r = cj 1, for j 4 and with no restric-
tions on the value of c (thus encompassing also r = mj 1), consider
d = E^a4* + a4(c)+2. Then kd = E^o" (49+1 + <*49+2) + EJLTa4 +
a4(m-c)+3 + 2a4(m-c+l) + a4(m-C+l)+2 Nqw 2a^m-c+l) = l SQ that d kd.
When r = cj, for j = 4 and c < m, consider d = a4c +
Q,4(m-c-l)+2 _|_ Q,4(rrec1)+3 _|_ Qf4(m-c)+2^ go that fcj ^^c-1 (a4ff+1 + a4ff+2) +
^m-c-1 ^4^ ^4(mc1)+3 _|_ gQ,4(m-c) _|_ ^(m-cj+l _|_ a4(m-c)+2 _|_ 2Q,4(m-c)+3
a4(m-of 1) + a4(m-C+l)+2_ Her6) 3a4(m-c) = a4(7n-c) + whjch doeg not oth-
86
erwise appear in the product unless c = 1, (where a4m + Y^=o a4ff+1 = a4m+1,
which does not otherwise appear), while 2a4^m_c^+3 = 1, so that d < kd.
In the case in which r = cj + 1 with no restrictions on either c or j (thus
encompassing r = mj + 1 as well), we use d Y=o + o^m~c^ + a.^m~c^+1,
so that kd = (9j+1 + 9j+2 + a(s+1)j) + a{m~c)j+1 + 2a(m_c>J'+2 +
Q(m-c)j+3 + a(m-c+l)j + a(m-C+l)j + l Since 2a(m-c)j+2 = ^ -< kd.
For r = mj 3 with j = 5, consider d = 1 + a + a5, where kd = a + 2a2 +
a3 + a5 + 2a6 + a7 + a10. Since 2a2 which cannot appear elsewhere,
and since 2a6 = 1, d < kd.
For r = mj 2 with j = 5, consider in similar fashion to the preceding case
d = 1 + a + ct3, where kd = a + 2a2 + a3 + a4 + 2a5 + a6 + a8. Here, 2a2 = amj,
which does not otherwise appear, and because 2a5 = 1, d ^ kd.
Now when j = 4 and r = mj, if m > 3, consider d 1 + a + a5 + a7, so
that kd = a + 2c? + ^c=3aC + 2a9 + a11. Now 2a2 = ami+2, which appears
nowhere else, while 2a9 = a6, 2a6 = a3, and 2a3 = 1, so that d -< kd. On the
other hand, when m = 2, k 23 + 26 + 27, so that = 23 + 26 = a + 7, so
that 3 (k) is a Glynn hyperoval in PG (2, 211).
Finally, when r mj + 2, set d = Y^=o a+ cdm-2^+2 + ajm-1b'+2 +
a(m~ 1b+4. In this case, kd = 2a + a2 + ]G(!!r(2 (aCJ + aCJ+1 + aC9+2) + o;(rn-2b+3 +
Q,(m-2p+4 _|_ a(m-i)j _|_ ac + a7"-74-2. Now 2a = 1, and if j = 4,
2Q;4m+2_|_^4m+i^2 _ 0,401-3^ does not otherwise appear, so that d < kd.
Thus, in all cases, either a d with d < kd or an equivalence to a Glynn
hyperoval was presented.
We finally summarize the results of the preceding propositions into a clas-
sification lemma.
Lemma 4.21 Ifk = a+oP+ai, where a 2J and 2 = ar, with h mj+2n+l,
then if @ (k) is a hyperoval, it is a translation hyperoval, a Segre hyperoval, or
a Glynn hyperoval.
Proof: Table 4.2 illustrates how Propositions 4.15-4.20 together classify this
case.
Table 4.2: Case 2 Classification Outline
Conditions Proposition
m > 2, n > 2 4.15
m = 1, j > 2n + 3 4.16
m = 1, j 2n + 3 4.17
m = 1, j = 2n + 2 4.18
n 0 4.19
n = 1 4.20
With all possibilities covered, the result follows.
4.3 Case 3
In case 3, where j = i + l 1 and h = j +1, we find it useful to deviate from
our usual technique of decomposing h as mj + ni + l and to instead decompose h
as ni + l'. The following proposition uses this idea to rule out many possibilities.
88
Proposition 4.22 Let k = a + cd + a?, where a = 2l, 2 = ar, h = j + l,
l + 2 < < 21 1, and j = i + l 1. Then (k) is not a hyperoval in
PG (2, 2h).
Proof: Consider d = X^=o_1 a+(^2l~1- Then kd = Y?l=l a +
a2i+i-i _j_ oca + a3l+l~2. Note that a3,+I_1 = al~l. We make several
observations regarding this product. Notice first of all that all terms of d are
present in kd, as 21 + i 1 = h and thus q;2/+*-1 = 1. Further notice that
21 i < l 1, since l + 1 < i; 21 < i + l 1, since l < i 1; 31 2 < h\ and
21 i < l 1, since again l < i 1. As such, no terms can possibly coincide,
and d ^ kd. m
We may further, with a little work, eliminate the possibility that l = i 1,
as the following proposition establishes.
Proposition 4.23 Let k = ot + a* + aj, where a 2l, 2 = ar, h = j + i 1
and j = 2i 2. Then Q) (k) is not a hyperoval in PG (2, 2h).
Proof: Let l = 2 1. We consider the congruence classes of l and r modulo
3 together with the value d = ^c=oa3c- Notice that h = 3i 3, so that
r ^ 0 mod 3, lest 3| (r, h). If l = 1 mod 3, kd = X^c=o (a3c+1 + 2a3c+2),
which reduces to either X^c=oa3c (when r = 2 mod 3) or (a3c + <*3c+1)
(when r = 1 mod 3), so that d < kd. If l = 0 mod 3 or l = 2 mod 3,
kd = X^c=o (a3c + 2a3c+1). Then, if r = 1 mod 3, kd = ]Cc=o (a3c + a3c+2) and
d ^ kd, so we need only consider r = 2 mod 3.
Similarly, however, consider d = ad. Here kd = (;d + 2ac,+ 1),
and unless r = l 1 mod l, d < kd. Thus, it remains only to consider the case
89
r = 2 mod 3, with l = 0 mod 3 or l = 2 mod 3, and r = l 1 mod l. It
follows immediately then that if l = 2 mod 3, the only remaining value for r is
r = 31 1, while if Z = 0 mod 3, we must also consider r = l 1 and r = 21 1.
Now if r = 3/ 1, let d = 1 + a1, so that Zed = 1 + a + 2al+1 + a2i + a2l+l.
As 2cd+1 = a1, d < kd.
If r = 21 1 and l > 6, consider d = X^c=o aC + a/+3 so that Zed = a +
a2 -f 2a3 + ^=]+i a + Y^2i a + 2/+4- Now 2a3 = a2?+2, 2a2l+2 = a/+1, and
2a/+1 = 1, so that d X Zed. If, on the other hand, i 3, set d = 1 + a + a5, so
that Zed = 1 + a + 2a2 + a4 + a5 + 2a6 + a7. Here, 2ct6 + 2a2 + a7 = a2 + a3,
so that d ^ Zed.
Finally, if r = l 1, consider d = X)c=o aC + cd+1 so that Zed = 2a + a2 + a3 +
al+l+2al+2+a/+3+a21+a2l+1+2a2/+2. Now 2a2l+2 = a, and 3a = a+al, which
does not otherwise appear, while 2ai+2 = a2/+1, and 2a2/+1 = 1, so that d < kd,
as desired, unless 1 = 3. If l = 3, we obtain additionally a 2a21 = a3*-1 = a2i+2
term, and we obtain a 2a; = ai+2 term. However, we still have d < kd.
Thus, a value of d with d ^ Zed was exhibited in all cases.
Our next series of results places restrictions on the values which r may
assume, restricting the value of l as little as possible.
Proposition 4.24 Let k = a + a1 + a-7, where a = 2l, 2 = aT, l ^ i 1,
j = i + l 1, and h = j + 1. If l + 2 < r <21 then @ (Ze) is not a hyperoval
in PG (2,2'*).
Proof: Consider the value d = l + ^Cc=r-z ftC- Then kd = 1 + a + ]Cc=t-m aC +
a1 + Ec=!+r-/ aC + 2aj + cd+1 + Y^cZ]+r-iaC- But 2qJ = aT~l-> so that d^kd.m
90
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