THREE-BIT MONOMIAL HYPEROVALS IN PG ( 2, 2 h)

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1 THREE-BIT MONOMIAL HYPEROVALS IN PG 2, 2 h) TIMOTHY L. VIS 1. Preliminaries Definition 1.1. In the projective plane PG 2, 2 h), we define the set D k) of 2 h +2 points as follows: D k) = { 1, x, x k) x GF 2 h)} {0, 1, 0), 0, 0, 1)}. We are interested in determining when this set of points determines a monomial hyperoval in PG 2, 2 h). The following definition and results of Glynn [2] will be useful to that end. Definition 1.2. Let a = h 1 i=0 a i2 i and b = h 1 b=0 b i2 i such that a i {0, 1}, b i {0, 1}, be the binary expansions of a and b. We say a b if and only if a i b i for all i. Theorem 1.3 Glynn s Criterion). D k) is a monomial hyperoval in PG 2, 2 h) if and only if there is no d such that 1 d 2 h 2 and d kd, when kd is reduced mod 2 h 1 to lie in [ 0, 2 h 1 ] with the convention that 0 is reduced to 0 and any other multiple of 2 h 1 is reduced to 2 h Reduction In order to classify those monomial hyperovals in PG 2, 2 h) with monomial exponent having binary expansion k = 2 i0 + 2 i1 + 2 i2, it is helpful to restrict our attention to a more limited class of exponents in an attempt to classify these hyperovals. Definition 2.1. Let k = n i=0, with a 2ai i < h. Further, let h = bc. If, for every pair i, j, a i a j mod b, we say that k is b-reducible with respect to h. The number k = n i=0 2a i, where a i is the reduction modulo b of a i to lie in [0, b 1], is called the b-reduction of k with respect to h. If k is not b-reducible, we say k is b-irreducible with respect to h. In a broader sense, we make the following definition. Definition 2.2. If there exists a b 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 b of h, we say that k is irreducible with respect to h. When we are working in a finite field of order 2 h, we say reducible and irreducible for reducible with respect to h and irreducible with respect to h respectively. This concept of reducibility allows a significant reduction in the scope of the necessary classification, as the following results will indicate. Our first result states Date: July 7,

2 2 TIMOTHY L. VIS that if a b-reduction of a value k does not determine a monomial hyperoval, k does not determine a monomial hyperoval. Theorem 2.3. Let h = bc and let k be a b-reduction of k with respect to h. If D k ) is not a monomial hyperoval in PG 2, 2 b), then D k) is not a monomial hyperoval in PG 2, 2 h). Proof. Suppose D k ) is not a monomial hyperoval in PG 2, 2 b). Then there exists some d such that 1 d 2 b 2 and d k d mod 2 b 1. Let d = c 1 i=0 2bi d. Now 1 d 2 h 1, so it remains only to show that d kd mod 2 h 1. Now 2 ji+a d 2 a d mod 2 h 1, so that kd k d mod 2 h 1. Thus, kd c 1 i=0 2bi k d mod 2 h 1. Since d is simply d repeated c times and now kd is simply k d repeated c times, we must have d kd, so that D k) is not a monomial hyperoval in PG 2, 2 h). 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 c-reduction of an exponent yielding a monomial hyperoval must itself give a monomial hyperoval. This suggests the following means of classification: 1) Classify all monomial hyperovals having irreducible exponents. 2) Use the classification of hyperovals having irreducible exponents to classify monomial hyperovals that have reducible exponents. In the case that k = 2 i0 + 2 i1 + 2 i2 we may easily restrict the occurrence of irreducible exponents. Theorem 2.4. If k = 2 i0 + 2 i1 + 2 i2 and p, q, and r are distinct primes dividing h, then k is reducible. Proof. Suppose h = p α q β s, where p s, q s, and s > 1. Further suppose that k = 2 i0 +2 i1 +2 i2 and that k is irreducible. Since k is irreducible, p α q β must divide one of i 2 i 1, i 1 i 0, and i 0 i 2. Without loss of generality, then, p α q β i 0 i 2. Again, since k is irreducible, p α s must divide one of the three differences. However, if p α s i 0 i 2, it must be that h i 0 i 2, a contradiction. Thus, p α s i 1 i 0 or p α s i 2 i 1. Without loss of generality, then, p α s i 1 i 0. But then, since p α i 0 i 2 and p α i 1 i 0, p α i 2 i 1. Finally, since k is irreducible, q β s must divide one of the three differences. However, if q β s i a i b, since p α i a i b, h i a i b. So q β s cannot divide any of the three differences, a contradiction, so that k must be reducible. 3. Common Divisor Conditions If k = 2 i0 + 2 i1 + 2 i2 it is helpful to the classification if we can normalize to one of the exponents. If, in fact, one of the exponents without loss of generality i 0 ) is relatively prime to h, a convenient normalization is possible, allowing us to adapt Glynn s criterion to a α-ary expansion, where α = 2 i0, since the α-ary expansion is simply a permutation of the indices on the binary expansion. Our main result is that for an irreducible exponent k = 2 i0 + 2 i1 + 2 i2, this is always possible. Our first few results, however, do not require that k be irreducible. Lemma 3.1. Suppose m > 1, and m divides each of i 0, i 1, i 2, and h. Then if k = 2 i0 + 2 i1 + 2 i2, D k) is not a hyperoval in PG 2, 2 h).

3 THREE-BIT MONOMIAL HYPEROVALS IN P G 2, 2 h 3 Proof. Let d = h m 1 j=0 2jm. Then kd h m 1 j=0 2 jm + 2 jm+1). Since jm + 1 am for any a, d kd and 1 d 2 h 2. Thus, D k) is not a hyperoval in PG 2, 2 h). We are able to further restrict the form of k with the following lemmas. Lemma 3.2. If i 0, h) > 1, i 1, h) > 1, i 2, h) > 1 and there exists no m dividing each of i 0, i 1, i 2 and h, then for some i j, there exists an n dividing i j and h, but not dividing i l for any l j. Proof. Since there is no m dividing each of the i j, the numbers i 0, h), i 1, h), and i 2, h) cannot all be equal. Choose a = max {i 0, h),i 1, h),i 2, h)}. If a occurs only once as i j, h), then n = a divides i j and h, but not i l for any l j. On the other hand, if a occurs twice, then let i j, h) a, so that either i j, h) a or i j, h) is a common divisor of each i l and h greater than 1. Since this is excluded by the hypothesis, i j, h) i l for l j, and n = i j ) satisfies the conditions of the lemma. We now assume, without loss of generality that a number m exists dividing i 0 and h but neither i 1 nor i 2. Lemma 3.3. If i 1 and i 2 are not both congruent to 1 mod m and k = 2 i0 +2 i1 + 2 i2, D k) is not a hyperoval in PG 2, 2 h). Proof. Let d = h m 1 j=0 2jm. Then kd = h m 1 ) j=0 2 jm + 2 jm+r1 + 2 jm+r2, where r1 and r 2 are the reductions of i 1 and i 2 mod m to lie in [0, m 1]. If r 1 and r 2 are not both m 1, 2 jm+r1 + 2 jm+r2 2 j+1)m, so that d kd and D k) is not a hyperoval in PG 2, 2 h). In the case that i 0, h), i 1, h), and i 2, h) are all distinct, we can strengthen this condition considerably. Lemma 3.4. Suppose i 0, h), i 1, h), and i 2, h) are distinct and each greater than one. Then i 1 i 2 1 mod i 0, h), i 0 i 2 1 mod i 1, h), and i 0 i 1 1 mod i 2, h) or D k) is not a hyperoval in PG 2, 2 h). Proof. Lemma 3.3 guarantees that, without loss of generality, i 1 i 2 1 mod i 0, h). Without loss of generality, i 1, h) does not divide i 2, h) and thus does not divide i 2. Further, since i 1 1 mod i 0, h), i 1, h) does not divide i 0, h) and thus does not divide i 0. So, applying Lemma 3.3 again, i 0 i 2 1 mod i 1, h). But then since i 2 1 mod i 0, h), i 2, h) does not divide i 0, and since i 2 1 mod i 1, h), i 2, h) does not divide i 1. So Lemma 3.3 now implies that i 0 i 1 1 mod i 2, h). We are left with two possibilities: either all three of i 0, h), i 1, h), and i 2, h) are distinct, or i 1, h) = i 2, h) 1 mod i 0, 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 D 2 i0 + 2 i1 ) [1]. Lemma 3.5. Let k = 2 i0 + 2 i1 + 2 i2, where 2 i 1, h) and 2 i 2, h) and where 2 i 0, h). Then D k) is not a hyperoval in PG 2, 2 h).

4 4 TIMOTHY L. VIS Proof. Notice that i 0 is necessarily odd, since i 1 is even and congruent to 1 mod i 0, h). Now let d = h 2 i=0 22i 1. Then kd = 2 h 2 i=0 22i 1 + h 2 i=0 22i. Combining these two sums yields kd = 2 h 2 i=0 2 2i = h 2 i=0 2 2i 1 = d, so that d kd, and D k ) is not a hyperoval in PG 2, 2 h). When q > 2, a little more work is needed. Our treatment of this case is adapted from similar arguments in Cherowitzo and Storme [1] dealing with k = 2 i + 2 j. Lemma 3.6. Let k = 2 i0 +2 i1 +2 i2 and let c i 1, h) and c i 2, h), where c > 1, with i 0 0 mod c. Now let i 1 = i1 c, i 2 = i2 c and h = h c. Further, let k = 2 i i 2. If there exists a d such that d k d and such that if 2 i 1 d = r I 1 2 r and 2 i 2 d = r I 2 2 r, I 1 I 2 = the products never coincide), then D k) is not a hyperoval in PG 2, 2 h). Proof. Where d = r D 2r, let d = r D 2cr. Now since 2 i 1 d = r I 1 2 r, then 2 i1 d = r I 1 2 cr. Similarly, if 2 i 2 d = r I 2 2 r, then 2 i2 d = r I 2 2 cr. Since I 1 I 2 = the products never coincide), all terms remain distinct, and since each r D is realized as 2 r for some r I 1 I 2, each cr with r D is realized as 2 cr for some r I 1 I 2. Finally, since d = r D 2cr and i 0 0 mod c, 2 i0 d = r I 0 2 r, where r 0 mod c. Thus, d kd, and D k) is not a hyperoval in PG 2, 2 h). We are immediately able to require that i 1, h) = i 2, h) = c. Lemma 3.7. Suppose k = 2 i0 + 2 i1 + 2 i2, and let c i 1, h) and c i 2, h), where i 1, h) i 2, h), and where c > 1. Further, let i 0 0 mod c. Then D k) is not a hyperoval in PG 2, 2 h) Proof. Let i 1 = i1 c, i 2 = i2 c, and h = h c. Then necessarily i 1, h ) = i1,h) c 1 and i 1, h ) i 2, h ) = i2,h) c. Then the proof of Lemma 3.7 in [1] provides a value for d satisfying the conditions of Lemma 3.6, so that D k) is not a hyperoval in PG 2, 2 h). The next lemma is an extremely close adaptation of Lemma 3.9 in [1]. Lemma 3.8. Suppose k = 2 i0 +2 i1 +2 i2, and let c = i 1, h),i 2, h)), where c > 1, and i 0 0 mod c. Let i 1 = ci 1, i 2 = ci 2, and h = ch. If D k) is a hyperoval in PG 2, 2 h), either i 2 2i 1 mod h or i 2 i 1 h +1 2 mod h, with h odd. Proof. If c i 1, h) or c i 2, h), Lemma 3.7 applies and D k) is not a hyperoval in PG 2, 2 h). Then we must have c = i 1, h) = i 2, h), so that i 1, h ) = 1. In this case, the proofs of Lemmas 3.4 and 3.5 in [1] provide values for d satisfying the conditions of Lemma 3.6 that eliminate every case except i 2 2i 1 mod h and i 2 i 1 h +1 2 mod h with h odd. We now consider the remaining possibilities. Since i 2, h ) = 1 as well, we must, in addition to having either i 2 2i 1 mod h or i 2 i 1 h +1 2 mod h with h odd, have either i 1 2i 2 mod h or i 1 i 2 h +1 2 mod h with h odd. Notice in particular, that if i 2 i 1 h +1 2 mod h, 2i 2 i 1 h + 1) i 1 mod h. As such, we have two possibilities: either i 1 2i 2 mod h with h odd, or i 1 2i 2 and i 2 2i 1.

5 THREE-BIT MONOMIAL HYPEROVALS IN P G 2, 2 h 5 The next two lemmas, closely adapted from Lemmas 3.11 and 3.12 in [1] rule out these cases. Lemma 3.9. Suppose k = 2 i0 + 2 i i1, where i 1, h) = 2i 1, h) = c > 1, and i 0, c) = 1, with h = h a odd. Then D k) is not a hyperoval in PG 2, 2 h). Proof. Let d = h 1 2 t=0. Then the following is obtained for kd: 22ti1 kd = h 1 2 t=0 = 2 i t+1)i1 + h 3 2 t=1 h +1 2 t=1 2 2ti t+1)i h 1 = i ti1 + t=2 h 1 2 t=0 h 1 2 t=0 h 1 2 t=1 2 2ti1+i0 2 2ti1 + 2 i ti1+i0 h 1 2 t=0 2 2ti1+i0 Since i mod c, while each ti 1 0 mod c, and since 2ti 1 + i 0 0 mod c the only terms which may possibly not be distinct are 2 i1+1 and 2 2ti1+i0. But if c > 2, even a possible 2 2 i1+1 2 mod c, so that the resulting terms are distinct and d kd. On the other hand, if c = 2, Lemma 3.5 yields the desired result, so that D k) is not a hyperoval in PG 2, 2 h). Lemma Let k = 2 i0 +2 i1 +2 i2, with i 1, h) = i 2, h) = c > 1 and i 0, c) = 1. Let h = h c, i 1 = i1 c, and i 2 = i2 c, and suppose that i 2 2i 1 mod h and that i 1 2i 2 mod h. Then D k) is not a hyperoval in PG 2, 2 h). Proof. Since i 2 2i 1 mod h and i 1 2i 2 mod h, then i 2 2i 1 mod h and i 1 2i 2 mod h, so that {i 1, i 2 } = { h 3, } 2h 3. But then h = 3, which is odd, and Lemma 3.9 applies, so that D k) is not a hyperoval in PG 2, 2 h). Theorem Let k = 2 i0 + 2 i1 + 2 i2, with i 0, h) > 1 and i 1, h) = i 2, h) > 1, where i 0 0 mod i 1, h). Then D k) is not a hyperoval in PG 2, 2 h). Proof. The conditions stated satisfy the hypotheses of Lemmas 3.5, 3.6, 3.8, 3.9, and 3.10, which leave no possibilities for k to describe a hyperoval. We note that, although the previous restrictions allowed us to assume that i 1, h) = i 2, h), this was not assumed for the initial lemmas and was, in fact, ruled out in Lemma 3.7, so that this series of lemmas in fact ruled out the existence of any m dividing i 1, i 2, and h, but not i 0. With the added condition that k be irreducible, we can, in fact enforce a further restriction. Lemma Suppose k = 2 i0 +2 i1 +2 i2 is irreducible and i 0, h) > 1, i 1, h) > 1, and i 2, h) > 1. Then {i 0, h),i 1, h),i 2, h)} = 2 or D k) is not a hyperoval in PG 2, 2 h). Proof. Lemma 3.1 guarantees that {i 0, h),i 1, h),i 2, h)} 1, so suppose that {i 0, h),i 1, h),i 2, h)} = 3. Then Lemma 3.4 guarantees that i 0 i 1 1 mod i 2, h), i 0 i 2 1 mod i 1, h), and i 1 i 2 1 mod i 0, h). These conditions ensure that any prime dividing i 0, h) does not divide i 1 or i 2, any prime

6 6 TIMOTHY L. VIS dividing i 1, h) does not divide i 0 or i 2, and any prime dividing i 2, h) does not divide i 0 or i 1. Thus, there are at least three distinct primes dividing h, so that k is reducible, contradicting Theorem 2.4. So {i 0, h),i 1, h),i 2, h)} = 2, as desired. With these preliminary steps in place, we are now prepared to prove the main theorem. Theorem Suppose k = 2 i0 +2 i1 +2 i2 is irreducible with i 0, h) > 1, i 1, h) > 1, and i 2, h) > 1. Then D k) is not a hyperoval in PG 2, 2 h). Proof. Lemma 3.12 ensures that under these conditions, exactly one of the greatest common divisors of i 0, i 1, and i 2 with h is distinct. Without loss of generality, assume i 0, h) i 1, h) = i 2, h). Further, by Theorem 2.4, h = p α q β. Lemma 3.3 guarantees that i 1 i 2 1 mod i 0, h), so that i 0, h) has no common factors with i 1 or i 2. Thus, without loss of generality, i 0, h) = p a and i 1, h) = q b, where 1 a α and 1 b β. Now let {i l, i m, i n } = {i 0, i 1, i 2 } in some order. Since k is irreducible, p α q β 1 divides one of the differences, without loss of generality i n i m. Further, p α 1 q β divides one of the differences, which cannot be i n i m lest h divide i n i m. Without loss of generality, then p α 1 q β i m i l. So p α 1 q β 1 divides both i n i m and i m i l, and therefore also divides i l i n. Thus, p α 1 q β 1 divides each of i 0 i 2, i 1 i 0, and i 2 i 1. Now define a = min {α 1, a} and b = min {β 1, b}. So p a divides each of i 0, i 0 i 2, and i 1 i 0, implying that p a also divides i 1, i 2, i 1, h), and i 2, h). Since i 1, h) = q β, a = 0, so that a = α = 1. Similarly, q b divides each of i 1, and i 1 i 0, implying that q b also divides i 0 and thus i 0, h). Since i 0, h) = p α, b = 0, so that b = β = 1. Thus, h = pq, q divides both i 1 and i 2, and i 1 i 2 1 mod p. Since the p multiples of q have distinct congruences modulo p, i 1 = i 2, a contradiction. So D k) is not a hyperoval in PG 2, 2 h). Thus, any monomial hyperoval in PG 2, 2 h) with irreducible exponent k = 2 i0 + 2 i1 + 2 i2 must satisfy i j, h) = 1 for some j. { 2 ) } i Note, however, that the sequence j i h 1 is merely a permutation of the sequence { 2 i} h 1 i=0. If then we let α =, we can express each integer in Z 2ij 2 h 1 in the form n = h 1 i=0 a iα i, where a i {0, 1} and the summation is reduced modulo 2 h 1 to lie in [ 0, 2 h 1 ] with the convention that 0 is reduced to 0 and any nonzero multiple of 2 h 1 to 2 h 1. Further, the a i 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 1.2, it is immediately evident that a b in the binary expansion if and only if a b in the expansion with respect to α. It follows then, that Theorem 1.3 applies equally well to this expansion. We may thereby restrict our attention to those exponents of the form α + α i + α j, and, in so doing, reduce the number of parameters to be considered by one. 4. Monomials With Exponent of the Form α + α i + α j 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 α = 2 ij, where i=0

7 THREE-BIT MONOMIAL HYPEROVALS IN P G 2, 2 h 7 i j, h) = 1. Using the division algorithm, we may express h as h = mj + r, where r < j. Further, since j is necessarily less than h, m 1. Again using the division algorithm, we may express r as r = ni + l, where l < i. Here, however, i need not be less than r, and n may be zero. Together, these two expansions allow us to express h as h = mj + ni + l, where ni + l < h 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 α + α i + α j. Lemma 4.1. Let k = α + α i + α j, where α = 2 i0 for i 0, h) = 1 and suppose that for some s > 1, s i, s j, and s h. Then D k) is not a hyperoval in PG 2, 2 h). Proof. Without loss of generality, suppose that α i = 2 i1 and α j = 2 i2. Then s i 1 and s i 2, but s i 0, and the results of Lemma 3.7 and Theorem 3.11 preclude D k) from being a hyperoval in PG 2, 2 h). Our next result shows that, in fact, once k can be expressed as α + α i + α j, one of i and j must be relatively prime to h, and therefore k can also be expressed as β + β x + β y, where β x = α and β = α i if i, h) = 1 or β = α j if j, h) = 1. Theorem 4.2. Let k = α + α i + α j, where α = 2 i0 and α r = 2. Suppose that i, h) = a and j, h) = b for a > 1 and b > 1. Then D k) is not a hyperoval in PG 2, 2 h). Proof. By Lemma 4.1, we may assume that a, b) = 1, so that ab h. If i 1 mod b, let d = h b 1 αcb, so that kd = h b 1 α cb + α cb+1 + α cb+i). No terms coincide as i 0 mod b since b i)and i 1 mod b. On the other hand, if i 1 mod b and r 1 mod b, kd = h b 1 α cb + α cb+r+1) and no terms coincide, so that d kd in either case. Similarly, if j 1 mod a, let d = h a 1 αca, so that kd = h a 1 α ca + α ca+1 + α ca+j). Again, no terms coincide as j 0 mod a since a j) and j 1 mod b. Once again, if j 1 mod a and r 1 mod a, kd = h a 1 α ca + α ca+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 = h ab 1 α cab + α cab+i 1), so that kd = h ab 1 α cab + α cab+1 + 2α cab+i + α cab+2i 1 + α cab+j). With r 1 mod ab, h ab 1 2α cab+i = h ab 1 α cab+i 1 and as no other terms coincide, d kd. On the other hand, if j 2i, let d = h ab 1 α cab + α cab+j 1), so that kd = h ab 1 α cab + α cab+1 + α cab+i + α cab+j i + 2α cab+j). Again, r 1 mod ab, so that h ab 1 2α cab+j = h ab 1 α cab+j 1 and as no other terms coincide, d kd. Thus, having covered all possible cases, if D k) is a hyperoval in PG 2, 2 h), either i, h) = 1 or j, h) = 1.

8 8 TIMOTHY L. VIS We now use five related values of d to place severe restrictions on the parameters of a potential hyperoval. Lemma 4.3. Let k = α + α i + α j, and let h = mj + ni + l, where ni + l < j and l < i. If D k) is a hyperoval in PG 2, 2 h), then at least one of the following conditions holds: 1) i 1 = l 2) j i ni + l j i + l 3) j 1 = ni + l. Proof. Suppose that none of the above conditions hold, and let d = l 1 n 1 b=0 αbi+l + m 1 αcj+ni+l. This leads to the following value for kd. l 1 kd = α a+1 + α a+i + α a+j) n 1 + α bi+l+1 + α bi+l+i + α bi+l+j) a=0 + m 1 b=0 α cj+ni+l+1 + α cj+ni+l+i + α cj+ni+l+j). a=0 αa + Under the conditions set out, d is defined. Further, all terms of d are present and no terms coincide, so that d kd and D k) is not a hyperoval in PG 2, 2 h). Lemma 4.4. Let k = α + α i + α j, and let h = mj + ni + l, where ni + l < j and l < i. If D k) is a hyperoval in PG 2, 2 h), then at least one of the following conditions holds: 1) n 1 2) i = 2 3) m = 1 and j {ni + l + 1, ni + l + i 1, ni + l + i} 4) l = 0. Proof. Suppose that none of the above conditions hold, and let d = l 1 a=0 αa + α i+l 1 + n 1 b=1 αbi+l + α j+ni+l i + m 1 c=1 αcj+ni+l. We then obtain the following expression for kd. l 1 kd = α a+1 + α a+i + α a+j) + α i+l + α 2i+l 1 + α j+i+l 1 a=0 n 1 + α bi+l+1 + α bi+i+l + α j+bi+l) b=1 + α j+ni+l i+1 + α j+ni+l + α 2j+ni+l i + m 1 c=1 α cj+ni+l+1 + α cj+ni+i+l + α cj+j+ni+l) Again, under the conditions imposed, d is defined, all terms of d are present, and no duplicate terms appear. Thus, d kd and D k) is not a hyperoval in PG 2, 2 h). Lemma 4.5. Let k = α + α i + α j, and let h = mj + ni + l, where ni + l < j and l < i. If D k) is a hyperoval in PG 2, 2 h), then at least one of the following conditions holds:

9 THREE-BIT MONOMIAL HYPEROVALS IN P G 2, 2 h 9 1) n = 0 2) i = 2 3) j = ni + l + 1 4) j i ni + l j + l i 1. 5) j = 2i 6) l = 0. Proof. Suppose that none of the above conditions hold and let d = l 1 a=0 αa + α i+l 1 + n 1 b=1 αbi+l + m 1 αcj+ni+l. We then obtain the following expression for kd. l 1 kd = α a+1 + α a+i + α a+j) + α i+l + α 2i+l 1 + α j+i+l 1 a=0 n 1 + α bi+l+1 + α bi+i+l + α j+bi+l) + b=1 m 1 α cj+ni+l+1 + α cj+ni+i+l + α cj+j+ni+l) Again, under the conditions imposed, d is defined, all terms of d appear, and no duplicate terms appear. Thus, d kd and D k) is not a hyperoval in PG 2, 2 h). Lemma 4.6. Let k = α + α i + α j, and let h = mj + ni + l, where ni + l < j and l < i. If D k) is a hyperoval in PG 2, 2 h), 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 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 = l 1 n 1 a=0 αa + b=0 αbi+l + α j+ni i+l + m 1 c=1 αcj+ni+l. We obtain the following expression for kd: l 1 kd = α a+1 + α a+i + α a+j) n 1 + α bi+l+1 + α bi+i+l + α j+bi+l) a=0 b=0 + α j+ni i+l+1 + α j+ni+l + α 2j+ni i+l + m 1 c=1 α cj+ni+l+1 + α cj+ni+i+l + α cj+j+ni+l) 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 D k) is not a hyperoval in PG 2, 2 h). A variation on this gives slightly different conditions.

10 10 TIMOTHY L. VIS Lemma 4.7. Let k = α + α i + α j, and let h = mj + ni + l, where ni + l < j and l < i. If D k) is a hyperoval in PG 2, 2 h), then at least one of the following conditions holds: 1) i = 2 2) j = i + 1 3) ni + i j ni + i + l 1 4) m = 1 5) l = 0. Proof. Suppose, as usual, that none of the above conditions hold, and let d = l 1 a=0 αa + n b=1 αbi+l 1 + α j+ni+l 1 + m 1 c=1 αcj+ni+l. We obtain the following expression for kd: l 1 kd = α a+1 + α a+i + α a+j) + a=0 n α bi+l + α bi+i+l 1 + α bi+j+l 1) b=1 + α j+ni+l + α j+ni+i+l 1 + α 2j+ni+l 1 + m 1 c=1 α cj+ni+l+1 + α cj+ni+i+l + α cj+j+ni+l) As before, d is defined, the terms of d appear, and no duplicate terms appear, so that d kd and D k) is not a hyperoval in PG 2, 2 h). Combining the conditions of Lemmas 4.1, 4.3, 4.4, 4.5, 4.6, and 4.7, we are able, after some analysis of different cases, to arrive at the following list of possible parameters for a monomial hyperoval of the form D α + α i + α j). As before, we express h as mj + ni + l, where ni + l < j and l < i. Theorem 4.8. Let k = α + α i + α j, where α = 2 i0 and α r = 2. Further let h = mj + ni + l, where ni + l < j and l < i. If D k) is a hyperoval in PG 2, 2 h), at least one of the following sets of conditions holds: 1) n = 1, m = 1, and 2i + 1 j 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) n = 0 and j i + l 10) i 1 = l, n = 0, and m = 1 11) i 1 = l, n = 1, and 2i j 3i 2 Proof. This follows from the results of Lemmas 4.1, 4.3, 4.4, 4.5, 4.6, and 4.7 by analyzing the possible combinations of conditions from each of the lemmas. We examine each of these cases in turn as the major portion of this classification.

11 THREE-BIT MONOMIAL HYPEROVALS IN P G 2, 2 h 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 α r = 2, with 1 r h 1. We use the following family {d a } of values for d, where l a d a = α b + b=0 i+l b=i+l a and 0 a l In this case, then, we obtain the following product kd a : kd a = l a+1 b=0 i+l+1 α b + b=i α b + 2i+l c=2i+l a α b 2i+x+l a α c + α c + c=2i+x 3i+x+l 1 b=3i+x+l a 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 powers of α that appear multiple times are precisely the powers α c, where max {2i + x, 2i + l a} c min {2i + l + x a, 2i + l}. Notice also that all terms of d a appear in kd a. The use of {d a } severely restricts the allowable values of r. Lemma 4.9. Let k = α+α i +α 2i+x, where α = 2 i0 with α r = 2 and 1 r h 1. Further, let h = 3i + l + x, where l < i and 1 x l. If min {x, l x} + 1 r i + x 2, 2i + l + x + 3 r 3i + l 1, or i + l + 3 r 2i + l 2, or if l i 1 and r {i + l + 2, 2i + l 1, 2i + l + x + 2, i + x 1}, then D k) is not a hyperoval in PG 2, 2 h). Proof. Given a value of a, we determine the values of r for which α c+r does not appear for any max {2i + x, 2i + l a} c min {2i + l + x a, 2i + l} and thus d a kd a. We consider each of the four possible combinations for the maximum and minimum terms separately. If 2i + x 2i + l a and 2i + l + x a 2i + l, the acceptable values of r are i + 2l a + 2 r 2i + x 1, and a satisfies l x a x. If 2i + x 2i + l a and 2i + l 2i + l + x a, the acceptable values of r are i + 2l a + 2 r 2i + a 1, 2i + 2l + 2 r 3i + l 1, and l x + 1 r i 1, and a satisfies max {x, l x} a l. If 2i + l a 2i + x and 2i + l + x a 2i + l, the acceptable values of r are i + l + x + 2 2i + x 1, 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 r 2i+a 1, 2i+l+x+a+2 3i+x+a 1, and a + 1 r i 1, and a satisfies x a l x. Considering at once all possible values of a in a given situation, the values i+l+ 3 r 2i+l 2 i+l+2 r 2i+l 1 if l < i 1), 2i+l+x+3 r 3i+l 1 2i + l + x + 2 r 3i + l 1 if l < i 1), and min {x, l x} + 1 r i + x 2 min {x, l x} + 1 r i + x 1 if l < i 1 each have some choice of a with 1 a l 1 0 a l if l < i 1) such that d a kd a, yielding the desired result. Lemma Let k = α + α i + α 2i+x, where α = 2 i0, α r = 2, and h = 3i + l + x, where l < i and 1 x l. If r min {x, l x} or r 3i + l, D k) is not a hyperoval in PG 2, 2 h) or it is a translation hyperoval, Segre hyperoval, or Glynn hyperoval. α b.

12 12 TIMOTHY L. VIS Proof. Consider the value kd 1. Notice that kd 1 has only the duplicate terms 2α 2i+l 1 and 2α 2i+l and no terms α c with 2i + x + l a + 1 c 3i + x + l a appear. Since this amounts to i consecutive terms and i > min {x, l x}, there exist q, q such that α 2i+l 1+qr and α 2i+l+q r do not otherwise appear in kd 1, while for s < q, s < q, α 2i+l 1+sr and α 2i+l+s r never appear in d 1, so that d 1 kd 1. In similar fashion, notice that r 3i + l is equivalent to h r x. No terms α c with i + l + 2 c min {2i + l 1, 2i + x} appear, amounting to i + x l 2 > x consecutive terms, except when l = i 1 and x i 2. Thus, unless l = i 1 and x i 2, using the same reasoning as above, d 1 kd 1. Now if l = i 1 and x = i 2, the only duplicate terms are 2α 3i 2 and 2α 3i 1, while if l = x = i 1, the only duplicate term is 2α 3i 1, and neither α 3i 2+r nor α 3i 1+r appears otherwise for r 3i+l+1. Thus, we have only to consider r = 3i+l in these two situations. If l = i 1 = x + 1 = h r 1, assume i > 3 if i = 2, Theorem 4.29 yields the result), and let d = 1 + α i 1 + α 2i 1 + α 3i 2, so that kd = 1 + α + α i 1 + 2α i + α 2i 1 + α 2i + α 3i 2 + 2α 3i 1. As 2α i = α 2 and 2α 3i 1 = α 2i+1, neither of which terms appear otherwise, d kd. If l = x = h r = i 1, let d = 1 + 2i 1 c=i+1 αc, so that kd = 1 + α + α i + 3i 2 c=i+2 αc + 2α 3i 1 + 5i 3 c=4i αc. Since 2α 3i 1 = α 2i and 2α 2i = α i+1, d kd, so that the result follows. The preceding lemmas now restrict the values of r in this case to i + x r i+l +1,and 2i+l r 2i+l+x+1 if l i 1, or to i+x 1 r i+l +2 and 2i + l 1 r 2i + l + x + 2 if l = i 1. We next examine these remaining cases, beginning with 2i+l 1 r 2i + l + x + 2. In this case we are immediately able to restrict r to the boundaries of this range. Lemma Let k = α+α i +α 2i+x, where α = 2 i0 with α r = 2 and 1 r h 1. Further, let h = 3i + l + x, where l < i and 1 x l. If 2i + l r 2i + l + x, then D k) is not a hyperoval in PG 2, 2 h). Proof. Choose d = i 2 b=0 αb + α 3i+l+x r 1. Since r 2i + x + l, all of these terms are distinct. Then kd = 2i 2 b=1 αb + 3i+x 2 b=2i+x αb + α 3i+l+x r + α 4i+l+x r 1 + α 5i+l+2x r 1. Since r 2i + l i + l + x + 1, α 3i+l+x r 1 appears in kd, and α 3i+l+x r appears exactly twice unless r = 2i + l and l = x = i 1. But 2α 3i+l+x r = 1, and if no further powers of α coincide, d kd. Since r 2i + l 2i + x, 5i + l + 2x r 1 3i + l + x, and α 5i+l+2x r 1 does not coincide with any other terms. Finally, since r 2i + l, 4i + l + x r 1 < 2i + x, so that α 4i+l+x r 1 does not coincide with any other terms. Thus, d kd, and D k) is not a hyperoval. In the case r = 2i + l and l = x = i 1, let d = i 1 αc + α i+l, so that kd = i 1 αc +2α i + 2i c=i+1 αc +2α 3i 1 + 4i 1 c=3i αc. Now 2α i +2α 3i 1 = α i +α 4i, and d kd again. This leaves r = 2i+l 1 when l = i 1), r = 2i+l+x+1, and r = 2i+l+x+2 when l = i 1) among the higher group of remaining values for r. Lemma Let k = α+α i +α 2i+x, where α = 2 i0 with α 2i+l+x+1 = 2. Further, let h = 3i + l + x, where l < i and 1 x l. Then D k) is not a hyperoval in PG 2, 2 h).

13 THREE-BIT MONOMIAL HYPEROVALS IN P G 2, 2 h 13 Proof. If l < i 2, let d = l b=0 αb + α i+l d 0 as defined earlier). Then kd = l+1 b=1 αb + i+l b=i αb + 2i+x+l b=2i+x αb +α i+l+1 +α 2i+l +α 3i+l+x. Note that α 3i+l+x = 1 and that the only power of α that appears twice is α 2i+l. But 2α 2i+l = α i+l+1. Since this already appears, we realize that 2α i+l+1 = α l+2, which cannot appear twice in the product kd. Thus, since all powers of α present in d are present in kd, d kd and D k) is not a hyperoval in PG 2, 2 h). On the other hand, if l i 2, let d = i 2 b=0 αb +α 2i 3. Then kd = 2i 3 b=1 αb + 2α 2i 3 + 3i+x 2 b=2i+x αb + α 3i 3 + α 4i+x 3. It may happen that a 2α 3i 3 term is present. But this is equal to α 2i 2 and we have either a 2α 2i 2 or 3α 2i 2 term present. In either case, this results in a second α i 1 term, which simplifies to a 1. Thus, d kd. Lemma Let k = α+α i +α 2i+x, where α = 2 i0 with α 2i+l+x+2 = 2. Further, let h = 3i + l + x, where l = i 1 and 1 x l. Then D k) is not a hyperoval in PG 2, 2 h). Proof. Let d = 1 + l b=2 αb + α i+l. Then kd = α + α i + α 2i+x + i i+l b=i+2 αb + 2i+x+l b=3 αb + b=2i+x+2 αb +α 2i +α 2i+l +α 3i+l+x. It may be that a 2α 2i+l term is present. But this is simply α 2i+1, which does not otherwise appear in the product, unless x = 1, in which case 2α 2i+1 = α i+1, which does not appear. In addition, a 2α i term is present, which simplifies to α 2, which is not otherwise present. Thus, all powers of α present in d are also present in kd, so that D k) is not a hyperoval in PG 2, 2 h). Lemma Let k = α + α i + α 2i+x, where α = 2 i0 with α 2i+l 1 = 2. Further, let h = 3i + l + x, where l = i 1 and 1 x l. Then D k) is not a hyperoval in PG 2, 2 h) unless it is a Segre hyperoval. Proof. Consider d l = 1 + i+l b=i αb as defined earlier. Then kd l = 1 b=0 αb + 2i b=i αb + α 2i+x + 2i+l c=2i αc + 3i+l+x 1 b=3i+x α b. The only powers that appear more than once are 2α 2i and 2α 2i+x. Now 2α 2i = α 4i+l 1 = α 3i+l+l = α l x, and so long as l x 2, this causes no further terms to coincide. On the other hand, 2α 2i+x = α 4i+l+x 1 = α i 1, which does not already appear. Since all terms of d are present in kd, d kd and the desired result is obtained. When l x = 1, k = h h 3. For i 3, let d = 2 αc, so that kd = 3 c=1 αc +α h 5 +α h 4 +2α h 3 +α h 2 +α h 1. As 2α h 3 +α h 2 +α h 1 = 1, d kd. If i = 2, k = and h = 7, so that 1 k = 6 and D k) is a Segre hyperoval. 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 α = 2 15s+1, α i = 2 10s+1, and α 3i 1 = 2 15s+2. In this case, set d = s s+1, so that kd = 2 5s s s s s s s+2, which simplifies to s s s s s+1, so that d kd. If i = 4s + 3, it is clear that h = 20s Further, it is easy to verify that α = 2 5s+2, α i = 2 10s+6, and α 3i 1 = 2 5s+3. In this case, set d = 1+2 5s s+7, so that kd = s s s s s m+10, which simplifies to s s s s+11, so that d kd.

14 14 TIMOTHY L. VIS We now address the situation in which i + x 1 r i + l + 2. We are again able to immediately restrict r to the boundaries of this range. Lemma Let k = α + α i + α 2i+x, where α = 2 i0 with α r = 2. Further, let h = 3i + l + x, where l < i and 1 x l. If i + x < r i + l, then D k) is not a hyperoval in PG 2, 2 h). Proof. Consider d = i+l+x r b=0 α b + α 2i+l+x r. Then kd = i+l+x r+1 b=1 α b + 2i+l+x r+1 b=i α b + 3i+l+2x r b=2i+x α b + α 3i+l+x r + α 4i+l+2x r. Notice that under the restrictions, 4i + l + 2x r < 3i + l + x and the only power of α that appears twice is α 3i+l+x r. But 2α 3i+l+x r = α 3i+l+x = 1, which does not otherwise occur. Thus, d kd and D k) is not a hyperoval in PG 2, 2 h). 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). Lemma Let k = α = α i + α 2i+x, where α = 2 i0 with α i+x = 2. If h = 3i + l + x, where l < i and 1 x l, then D k) is not a hyperoval in PG 2, 2 h). Proof. Let d = i 2 b=0 αb + α i+l x 1. Then kd = 2i 2 b=1 αb + α i+l x + α 2i+l x + 3i+x 2 b=2i+x αb + α 3i+l. Now 2α i+l x = α 2i+l appears, yielding 2α 2i+l = 1. It is possible that 2α 2i+l x = α 3i+l appears. In this case, 2α 3i+l = α i, 2α i = α 2i+x, and 2α 2i+x = α 3i+2x, which does not otherwise appear unless l = x, in which case, a second α 2i+x no longer appeared. So d kd. Lemma Let k = α + α i + α 2i+x, where α = 2 i0 with α i+x 1 = 2. Further, let h = 3i + l + x, where l < i and 1 x l. Then D k) is not a hyperoval in PG 2, 2 h). Proof. Set d = l x+1 b=0 α b + α 2i+l. Then kd = l x+2 b=1 α b + 2α i + i+l x+1 b=i+1 α b + 2i+l b=2i+x αb + 2α 2i+l+1 + α 3i+l. We notice that the only doubled powers of α are 2α i or 3α i when x = 1) and 2α 2i+l+1. Now 2α i = α 2i+x 1, which does not occur elsewhere in the product. Further, 2α 2i+l+1 = α 3i+l+x=1, so that d kd and D k) is not a hyperoval in PG 2, 2 h). Lemma Let k = α + α i + α 2i+x, where α = 2 i0 with α i+l+1 = 2. Further, let h = 3i + l + x, where l < i 2 and 1 x l. Then D k) is not a hyperoval in PG 2, 2 h). Proof. Consider d l = 1 + i+l b=i αb. Then kd = α + α i + α 2i+x + i+l+1 b=i+1 αb + 2i+l b=2i αb + 3i+l+x b=3i+x αb. Since α 3i+l+x = 1, all terms of d are present in the product. In addition, a 2α 2i+x appears. But 2α 2i+x = α, and 2α = α i+l+2. Since l < i 2, α i+l+2 does not otherwise appear in the product, and thus d kd. Lemma Let k = α + α i + α 2i+x, where α = 2 i0 with α i+l+1 = 2. Further, let h = 3i + l + x, where l = i 2 and 1 x l. Then D k) is not a hyperoval in PG 2, 2 h). Proof. Set d = x 1 b=0 αb + α i+x 1 + α 2i+2x 2. Then kd = x b=1 αb + i+x 1 b=i α b + 2i+2x 1 b=2i+x αb + α i+x + α 2i+x 1 + α 3i+2x 1 + α 2i+2x 1 + α 3i+2x 2 + α 4i+3x 2. Now α 4i+3x 2 = α 2x. It is possible that a 2α 2x term exists, however, no α 2x can be present in d. If a 2α 2x term exists, it simplifies since l = i 2) to α 2i+2x 1.

15 THREE-BIT MONOMIAL HYPEROVALS IN P G 2, 2 h 15 As such, either a 3α 2i+2x 1 term or a 2α 2i+2x 1 term exists. In either case, an additional α x term is produced. But 2α x = α 2i+x 1, and the resulting 2α 2i+x 1 term reduces to 1. All powers of α present in d being retained, d kd and D k) is not a hyperoval in PG 2, 2 h). Where l = i 1 rather than i 2, a slight adjustment to the value of d eliminates most of the remaining possibilities. Lemma Let k = α + α i + α 2i+x, where α = 2 i0 with α i+l+1 = 2. Further, let h = 3i + l + x, where l = i 2 and 1 x l. Then D k) is not a hyperoval in PG 2, 2 h). Proof. We consider three cases on the values of x. If x 3, we consider d = x 2 b=0 αb + i+x 1 b=i+x 2 αb +α 2i+2x 3. In this case, kd = x 1 b=1 αb + i+x b=i αb +α 2x 2 + 2i+2x 3 b=2i+x 2 αb +2α 2i+2x 2 + 3i+2x 1 b=3i+2x 3 αb. It is possible that a 2α 2x 2 = α 2i+2x 2 term exists, so either a 3α 2i+2x 2 term or a 2α 2i+2x 2 term exists. In either case, an additional α x 1 term is produced. But 2α x 1 = α 2i+x 1, and the resulting 2α 2i+x 1 term reduces to 1. All powers of α present in d being present in kd, with no further terms coinciding, d kd. If x = 2, set d = 2 b=0 αbi. Then kd = 2 b=0 αbi b=1 αbi + 4 b=2 αbi+2. Now α 4i+2 = α, so that a 2α appears. But 2α = α 2i+1, and 2α 2i+1 = 1, so that d kd in this case. Finally, if x = 1, h = 4i and r = 2i. Thus, this case cannot occur, since h, r) > 1. Thus, D k) is not a hyperoval in PG 2, 2 h). All that remains now is r = i + l + 2 with l = i 1. Lemma Let k = α + α i + α 2i+x, where α = 2 i0 with α i+l+2 = 2. Further, let h = 3i + l + x, where l = i 1 and 1 x l. Then D k) is not a hyperoval in PG 2, 2 h). Proof. We consider five cases on the values of x. If x 5, let d = x 4 i+x 1 b=i+x 4 αb +α 2i+2x 6. Then kd = x 3 b=1 αb + i+x b=i αb +α 2x 5 + 2i+2x 4 b=0 αb + b=2i+x 4 αb + α 2i+2x 5 + α 3i+2x 6 + 3i+2x 1 b=3i+2x 4 αb. It is possible that a 2α 2x 5 = α 2i+2x 4 term appears, in which case 2α 2i+2x 4 = α x 2, which cannot otherwise appear. Necessarily, however, a 2α 2i+2x 5 = α x 3 appears, and 2α x 3 = α 2i+x 2 and 2α 2i+x 2 = 1 so that d kd. If now x = 4 consider d = 2 b=0 αbi +α 2i+1. Then kd = 2 b=0 αbi b=2 αbi+4 +α 2i+2 +α 3i+1 +α 4i+5. Now α 4i+4 = α, so that a 2α = α 2i+2 appears. Now 2α 2i+2 = 1, so that d kd in this case. When x = 3, h = 4i + 2 and r = 2i + 1, so that h, r) > 1, and this case does not occur. When x = 2, consider d l 1. Now 2α 2i+2 + 2α 2i+3 appears in kd l 1 ; however, 2α 2i+2 + 2α 2i+3 + α 2 = α 2i+3 + α 3, neither of which terms otherwise appear, so that d l 1 kd l 1. Finally, when x = 1, let d = i 1 b=0 αb + α 2i 1. Then kd = i 1 b=0 αb + 2α i + 3i 2 b=i+1 αb + 2α 3i 1 + α 3i. Now 2α 3i 1 + 2α i = α i + α 3i+1 so that d kd. Thus, D k) is not a hyperoval in PG 2, 2 h). b=1 αbi +

16 16 TIMOTHY L. VIS Theorem Let k = α + α i + α 2i+x, where α = 2 i0 with i 0, h) = 1. Further, let h = 3i + l + x, where l < i and 1 x l. Then D k) is not a hyperoval in PG 2, 2 h). Proof. Lemmas 4.9 through 4.21 eliminated all possibilities for D k) to be a hyperoval in PG 2, 2 h) 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 lemma states that m and n may not simultaneously be greater than one. Lemma Let k = α + α 2 + α j, where α r = 2. Further, let h = mj + 2n + 1, where 2n+1 < j. If m 2 and n 2, then D k) is not a hyperoval in PG 2, 2 h). Proof. Define the family {d a,b } of values for d in the following way. If a n b, d a,b = a α2c + n b c=a α2c+1 + n c=n b αj+2c+1 + m 1 c=2 αcj+2n+1. If, on the other hand, a n b, d a,b = n b α2c + a c=n b αj+2c + n c=a αj+2c+1 + m 1 c=2 αcj+2n+1. Note that d n 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 n b,b, and when we wish to use the value defined under the a n b assumption, we will denote it by d + n b,b to avoid confusion. We now determine the value kd a,b for each of the definitions of d a,b. If a n b, we find the following value for kd a,b. kd a,b = 2a n b)+3 α c + 2α 2a+2 + j+2n+1)+1 c=j+2n b)+1 m 1 c=2 α c + c=2a+3 n 1 c=n b α c + α 2j+2c+1 a α cj+2n+1 + α cj+2n+2 + α cj+2n+3). n b 1 α j+2c + c=a α j+2c+1 The exponents which appear are distinct whenever 1 b n 1. Every power of α present in d a,b is present in kd a,b, and the only duplicate term which appears is 2α 2a+2. In a similar fashion, we compute the value kd a,b for the case in which a n b. In this case, we obtain the following: kd a,b = 2n b+1) + a c=n b n b 1 α c + n 1 α 2j+2c + c=a α j+2c + α 2j+2c+1 + j+2a+1 c=j+2n b) m 1 c=2 α c + 2α j+2a+2 + j+2a+3 c=j+2a+3 α c α cj+2n+1 + α cj+2n+2 + α cj+2n+3). Once again, all of the exponents are distinct so long as 1 b n 1), all powers of α that appear in d a,b also appear in kd a,b, and the only duplicate term which appears is 2α j+2a+2. We can now determine the values of r that are not eliminated by each of d 0,b, d n b,b, d+ n b,b, and d n,b.

17 THREE-BIT MONOMIAL HYPEROVALS IN P G 2, 2 h 17 For d 0,b the values that remain are 1 r 2 n b) 1 r = cj + 2n 1 : 2 c m 1 r = j 2 r = cj + 2n : 2 c m 1 r = j + 2c 1 : 0 c n b 1 r = cj + 2n + 1 : 2 c m 1 j + 2 n b) 1 r j + 2n + 1 mj + 2n 1 r mj + 2n r = 2j + 2c 1 : n b c n 1. For d n b,b the values that remain are r = 1 r = cj + 2b 1 : 2 c m 1 r = j 2c : 1 c n b + 1 r = cj + 2b : 2 c m 1 j 1 r j + 2b + 1 r = cj + 2b + 1 : 2 c m 1 r = 2j + 2c 1 : 0 c b 1 mj + 2b 1 r mj + 2n. For d + n b,b the values that remain are 1 r 2b + 1 r = cj + 2b + 1 : 1 c m 2 r = j 2 m 1)j + 2b 1 r m 1)j + 2n + 1 r = j + 2c 1 : 0 c b 1 r = mj + 2c 1 : b c n 1 r = cj + 2b 1 : 1 c m 2 mj + 2n 1 r mj + 2n r = cj + 2b : 1 c m 2. For d n,b the values that remain are r = 1 r = cj + 1 : 1 c m 2 r = j 2c : 1 c b + 1 m 1)j 1 r m 1)j + 2 n b) + 1 r = cj 1 : 1 c m 2 r = mj + 2c 1 : 0 c n b 1 r = cj : 1 c m 2 mj + 2 n b) 1 r mj + 2n. Under the assumption that 1 b n 1, examining the remaining values for these values of d a,b reveals that only the following values of r are not eliminated by at least one of the d a,b. r = 1 r = 2j + 1 r = j 2 r = m 1)j + 2b + 1 r = j 1 r = mj + 2n 1 r = j + 1 r = mj + 2n. We eliminate these values of r in turn. For r = 1, the only duplicate term in kd n,1 is 2α j+2n+2 = α j+2n+3, and 2α j+2n+3 = α j+2n+4, which does not otherwise appear when j 2n + 2. When j = 2n + 2, 2α j+2n+4 = α j+2n+5, which does not appear, so that d n,1 kd n,1. If r = j 2 and j 2n+3, the only duplicate term in kd n,0 is 2α2n+2 or 3α 2n+2, and 2α 2n+2 = α 2j+2n 2, which does not otherwise appear, so that d n,0 kd n,0. When j = 2n + 3, let d = n+1 α2c + m 1 c=1 αcj+2n+1. Then kd = 2n+2 αc + 2α 2n+3 +α 2n+4 + n c=1 αj+2c +2α j+2n+2 +α j+2n+3 + m 1 c=2 α cj+2n+1 +α cj+2n+2 + α cj+2n+3). Now 2α 2n+3 = α j+2n+1, so all terms of d appear in kd, and 2α j+2n+2 = α 2j+2n, which does not otherwise appear so that d kd. Now if r = j 1 and j > 2n + 3, the only duplicate term in kd 0,0 is 2α 2 = α j+1 term. Now 2α j+1 = α 2j, which cannot otherwise appear in kd 0,0, so that d 0,0 kd 0,0. If j = 2n + 3, let d = n+1 α2c + m 1 c=2 αcj+2n+1. Then kd = 2n+2 c=1 αc + 2α 2n+3 + α 2n+4 + n+1 c=1 αj+2c + m 1 c=2 α cj+2n+2 + α cj+2n+3) + m c=3 αcj+2n+1.

18 18 TIMOTHY L. VIS As 2α 2n+3 = α j+2n+2 and 2α j+2n+2 = α 2j+2n+1, d kd. Finally, if j = 2n +2, let d = n α2c + m 1 c=1 αcj+2n+1, so that kd = 2n+1 αc + 2α 2n+2 + n c=1 αj+2c + α j+2n+2 + α j+2n+3 + m 1 c=2 α cj+2n+1 + α cj+2n+2 + α cj+2n+3). Since 2α 2n+2 = α j+2n+1, d kd. If r = j+1, consider the value d 0,1. Since 2α 2 = α j+3 and 2α j+3 = α 2j+4, which cannot otherwise appear, d 0,1 kd 0,1. c=2 αcj+2n+1, so When r = 2j + 1, take d = n 1 α2c + α 2n 1 + α j+2n 1 + m 1 that kd = 2n 1 c=1 αc +2α 2n + n 1 αj+2c +α j+2n 1 +α j+2n +α j+2n+1 +α 2j+2n 1 + m 1 c=2 α cj+2n+2 + α cj+2n+3) + m c=3 αcj+2n+1. Since 2α 2n = α 2j+2n+1, d kd. For r = m 1)j + 2n 1, let d = 1 + α + α m 1)j+2n+1. Then kd = 1 + α + 2α 2 + α 3 + α j + α j+1 + α m 1)j+2n+2 + α m 1)j+2n+3. But 2α 2 = α m 1)j+2n+1 so that d kd. If r = mj + 2n 1, take d = 1 + α. Then kd = α + 2α 2 + α 3 + α j + α j+1, and 2α 2 = 1, so that again d kd. Finally, when r = mj +2n, if j > 2n+4 set d = n+1 α2c + m 1 c=1 αcj+2n+1. So α cj+2n+1 +α cj+2n+2 + kd = 2n+3 αc + n αj+2c +2α j+2n+2 +α j+2n+3 + m 1 c=2 α cj+2n+3). Now 2α j+2n+2 = α j+2n+1, so that d kd. On the other hand, if j 2n+4, set d = 1 + α 2n+2 + m 1 c=1 αcj+2n+1 so that kd = 1 + α +α 2 + α 2n+3 + α 2n+4 +α j +2α j+2n+2 +α j+2n+3 + m 1 c=2 α cj+2n+1 + α cj+2n+2 + α cj+2n+3). Now α j { α 2n+2, α 2n+3, α 2n+4}. As 2α 2n+4 = α 2n+3 and 2α 2n+3 = α 2n+2, and α 2n+2 does not appear apart from α j, α 2n+2 appears in each case. Further, 2α j+2n+2 = α j+2n+1, so that d 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, D k) is not a hyperoval in PG 2, 2 h). We next consider the situation in which m = 1 and j > 2n + 3. Lemma Let k = α + α 2 + α j, where α = 2 i0, α r = 2, j > 2n + 3, and h = j + 2n + 1. Then if D k) is a hyperoval in PG 2, 2 h), it is a translation hyperoval. Proof. As defined in the proof of Lemma 4.23, consider d a,0. Note now that kd a,0 can be simplified to 2a+1 αc +2α 2a+2 + 2n+3 c=2a+3 αc + a αj+2c + n 1 αj+2c+1. Notice also that only the following values of r are not ruled out for a particular a: 1 r 2 n a) + 1 r = j + 2c 1 : 0 c n a 1 r = j 2c : 1 c a + 1 j + 2 n a) 1 c j + 2n. For a = 0, this equates to the following values of r: 1 r 2n + 1 r = j + 2c 1 : 0 c n 1 r = j 2 j + 2n 1 r j + 2n while for a = n, this leaves the following values of r: r = 1 r = j 2c : 1 c n + 1. j 1 r j + 2n

19 THREE-BIT MONOMIAL HYPEROVALS IN P G 2, 2 h 19 Combining these two lists, only the following values of r remain noting that the lists may, in fact, be identical if n = 0): r = 1 r = j + 2c 1 : 0 c n 1 r = j 2c : 2 j 2n 1 2 c n + 1 j + 2n 1 r j + 2n r = j 2. We again rule out each of the remaining cases in turn. When r = 1, consider d n,0. Since 2α 2n+2 = α 2n+3 and 2α 2n+3 = α 2n+4 which may not otherwise appear in the product) and 2α 2n+4 = α 2n+5, where α 2n+5 does not appear in the product kd n,0 otherwise, d n,0 kd n,0. This fails only if n = 0 and j = 4, in which case, k = , and k = 22, so that D k) is a translation hyperoval in PG2, 4). For r = j 2c, we consider two cases depending on the parity of j. If j is even, consider d 0,0. In the product kd 0,0 a 2α 2 appears. Now with j even, r is also even. But every power of α greater than 2 in d is odd. Furthermore, h is odd, and the powers α j+2n 2c for 0 c n 1 are all absent from the product kd 0,0 Since this r value is necessarily less than 2n+1, if new powers of α continue to arise from the 2α 2 through continued carrying, one of these powers must be of the form α j+2n 2c with 0 c n 1, so that d 0,0 kd 0,0. This again fails for n = 0, however, n = 0 implies that j 1 2, so that j = 3, h = 4, and 1 k = 2, so that D k) is a hyperconic. If j is odd, and j 2c < 2n + 1, we use d = n+c j+1 2 g=0 α 2g + α j+1 2n+c 2 )+1 + α 2n+1. Then kd = 2n+2c j g=0 α g + 2α 2n+2c j+1 + α 2n+2c j+2 + α 2n+2 + α 2n+3 + n+c j+1 2 j+2n+c j+1 g=0 α j+2g +α 2 )+1, with all of the powers distinct. But 2α 2n+2c j+1 = α n+1, so that d kd. On the other hand, if j 2c = 2n + 1, we divide into two further cases. If j > 4n + 4, consider d 0,0. Then kd has a 2α 2 term. Now 2α 2 = α 2n+3 and 2α 2n+3 = α 4n+4. Since 4n + 4 < j, this does not otherwise appear, so that d 0,0 kd 0,0. If j < 6n + 3, consider d = 1 + α h α h 2 ) h 4 n 1 g=0 α 2g. Here, kd = ) 2 h 4 2n g=1 α g + h 4 n 1 g=0 α j h 2 +2g ). Notice that with j < 6n+3, all of the exponents in the second factor are less than h 2. Now j h 2 is equal to either 2 h 4 2n 1 or 2 h 4 2n, and in either case, 2α j h 2 = α h 2, 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 + n,0. In the product kd n,0, 2α 2n+2 = α j+2n, and 2α j+2n = α j 3 which does not appear, so that d + n,0 kd+ n,0. If j = 2n + 6, let d = 1 = n α2c+3 so that kd = 1 + α + 2α 2 + j c=4 αc + n 2 αj+2c+3. Now 2α 2 = α j and 2α j = α 3, so that k k 1 d kd. This fails when n = 0, in which case = 6 and D k) is a Segre hyperoval. When j = 2n + 5, h = 4n + 6 and r = 2n + 3 = h 2, so that r, h) 1, a contradiction. Finally, for j = 2n + 4 consider d = n+1 α2c. Finally, if j = 2n + 4 let d = n+1 α2c. Then kd = 2α + j 1 c=3 αc + 2α j + n c=1 α2c+j. But 2α j = α, 3α = α + α j 1, and 2α j 1 = 1, so that d kd.

20 20 TIMOTHY L. VIS When r = j + 2c 1 with 1 c n 1, we use d = n c g=0 α2c + α 2n c)+1, so that kd = 2n c)+1 g=1 α g + 2α 2n c)+2 + α 2n c)+3 + n c g=0 αj+2g + α j+2n c)+1. Now 2α 2n c)+2 = 1, so that d kd. If r = j 1, we consider several cases. If j > 2n + 6, consider d n 1,0 note that with 0 c n 1, n is necessarily greater than 0). In the product kd n 1,0, a 2α 2n appears. Now 2α 2n = α j+2n 1 and 2α j+2n 1 = α j 3. Since j > 2n + 6, j 3 > 2n + 3, so that d n 1,0 kd n 1,0. Now, if j = 2n + 6 and n > 1, set d = n 2 α2c +α 2n 3 +α 2n+1, so that kd = 2n 3 αc +2α 2n 2 +α 2n 1 +α 2n+2 + α 2n+3 + n 2 αj+2c + α j+2n 3. Now 2α 2n 2 = α j+2n 3 and 2α j+2n 3 = α 2n+1, so that d kd. If j = 2n + 6 and n = 1, set d = 1 + α 2 + α 6, so that kd = α+α 2 +2α 3 +α 4 +α 7 +2α 8 +α 10. But 2α 3 = α 10 and 2α 10 = α 6, while 2α 8 = α 4 and 2α 4 = 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 + n+1 c=1 α2c+1, so that kd = 1 + α + 2α 2 + 2n+3 c=4 αc + 2α 2n+4 + α 2n+5 + n 1 c=1 αj+2c+1. Now 2α 2n+4 = α 2, 3α 2 = α 2n+5 + α 2, and 2α 2n+5 = α 3, so that d kd. We next consider r = j + 2n 1. As in the proof of Lemma 4.23, we use d = 1 + α, so that kd = α + 2α 2 + α 3 + α j + α j+1. So long as n 0, α j+1 1, in which case 2α 2 = 1 and d kd. If n = 0, consider d = 1 + α 2 + α 3. Then kd = 2α + 2α 2 + α 3 + 2α 4 + α 5 + α j. Now 2α 4 = α 2, 3α 2 = α 2 + 1; 2α = α j, and 2α j = α j 2, which no longer otherwise appears unless j {3, 4, 5, 7}. If j = 3, 1 k = 2 and D k) is a hyperconic; if j = 4, 1 k = 6 and D k) is a Segre hyperoval; if j = 5 or j = 7, r, h) = 2 and D k) is not a hyperoval. Finally, if r = j + 2n, the values of d used in the proof of Lemma 4.23 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 D k) was shown to be a translation hyperoval. We next deal with the situation in which j = 2n + 3. In such a situation, we have the following result. Lemma Let k = α + α 2 + α j, where α = 2 i0, α r = 2, j = 2n + 3, and h = j + 2n + 1. Then if D k) is a hyperoval in PG 2, 2 h) it is a hyperconic. Proof. We split into cases for different values of r. Note that h = 4 n + 1), so we need only consider odd values of r. If r = 1, then consider d n,0. A 2α2n+2 term appears, creating a 3α 2n+3 term, which in turn creates an α 2n+4 term, which does not otherwise appear unless n = 0, in which case j = 3, and 1 α + α 2 + α 3) = α. But α = 2, so that D α + α 2 + α 3) is a hyperconic in PG 2, 2 4). Thus, for n > 0, d n,0 kd n,0. Now, if 3 r 2n 1, consider d = 1 + α + n c= r+1 α 2c+1. Then kd = 2 1+α+2α 2 +α 3 + 2n+2 c=r+3 αc +2α 2n+3 +α 2n+4 + n 1 α j+2c+1. But 2α 2 = α r+2, c= r+1 2 while 2α 2n+3 = α 2n+r+3, which does not otherwise appear, so d kd. When r = 2n + 1, we set d = 1 + α, so that kd = α + 2α 2 + α 3 + α 2n+3 + α 2n+4, and 2α 2 = α 2n+3, giving rise to a 2α 2n+3 = 1, so that d kd. On the other hand, for r = 2n + 3, we use d 0,0. Then 2α 2n+3 = α 2, and 3α 2 = α 2 + α 2n+5, which does not otherwise appear unless n = 0, in which case, k = α + α 2 + α 3 and h = 4, so that k = , and 1 k = 2, making D k) a hyperconic in PG 2, 2 4). For n > 0, then, d 0,0 kd 0,0.