ON THE NUMBER OF QUADRATIC ORTHOMORPHISMS THAT PRODUCE MAXIMALLY NONASSOCIATIVE QUASIGROUPS

Abstract Let q be an odd prime power and suppose that 
$a,b\in \mathbb {F}_q$
 are such that 
$ab$
 and 
$(1{-}a)(1{-}b)$
 are nonzero squares. Let 
$Q_{a,b} = (\mathbb {F}_q,*)$
 be the quasigroup in which the operation is defined by 
$u*v=u+a(v{-}u)$
 if 
$v-u$
 is a square, and 
$u*v=u+b(v{-}u)$
 if 
$v-u$
 is a nonsquare. This quasigroup is called maximally nonassociative if it satisfies 
$x*(y*z) = (x*y)*z \Leftrightarrow x=y=z$
 . Denote by 
$\sigma (q)$
 the number of 
$(a,b)$
 for which 
$Q_{a,b}$
 is maximally nonassociative. We show that there exist constants 
$\alpha \approx 0.029\,08$
 and 
$\beta \approx 0.012\,59$
 such that if 
$q\equiv 1 \bmod 4$
 , then 
$\lim \sigma (q)/q^2 = \alpha $
 , and if 
$q \equiv 3 \bmod 4$
 , then 
$\lim \sigma (q)/q^2 = \beta $
 .


Introduction
A quasigroup (Q, * ) is a nonempty set Q with a binary operation * such that, for each a, b ∈ Q, there exist unique x, y ∈ Q for which a * x = b and y * a = b.A quasigroup (Q, * ) is said to be maximally nonassociative if holds for all u, v, w ∈ Q.By [11], a maximally nonassociative quasigroup has to be idempotent (that is, u * u = u for all u ∈ Q).Hence, in a maximally nonassociative quasigroup, the converse of implication (1-1) holds as well.
The existence of maximally nonassociative quasigroups was an open question for quite a long time [4,10,11].In 2018, a maximally nonassociative quasigroup of order nine was found [5], and that was the first step to realise that Stein's nearfield construction [14] can be used to obtain maximally nonassociative quasigroups of all 2 A. Drápal and I. M. Wanless [2] orders q 2 , where q is an odd prime power [3].A recent result of the present authors [6] (partially duplicated in [13]) constructs examples of all orders with the exception of a handful of small cases and two sparse subfamilies within the case n ≡ 2 mod 4.
The main construction used in [6,13] is based upon quadratic orthomorphisms and applies for all odd prime powers q 13.However, it was left open how many quadratic orthomorphisms can be used in the construction.We provide an asymptotic answer to that question in this paper.Throughout this paper, q is an odd prime power and F = F q is a field of order q.For a, b ∈ F, define a binary operation on F by This operation yields a quasigroup if and only if both ab and (1 − a)(1 − b) are squares, and both a and b are distinct from 0 and 1, see [7,16].Denote by Σ = Σ(F) the set of all such (a, b) ∈ F × F for which a b.For each (a, b) ∈ Σ, denote the quasigroup (F, * ) by If a = b ∈ F \ {0, 1}, then Equation (1-2) defines a quasigroup in which u * (v * u) = (u * v) * u for all u, v ∈ F. This means that such a quasigroup is never maximally nonassociative.If q 13, then there always exists (a, b) ∈ Σ(F q ) such that Q a,b is maximally nonassociative [6,13].This paper is concerned with the density of such (a, b).Our main result is the following theorem.THEOREM 1.1.For an odd prime power q, denote by σ(q) the number of (a, b) ∈ Σ(F q ) for which Q a,b is maximally nonassociative.Then lim q→∞ σ(q) q 2 = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 953 • 2 −15 ≈ 0.029 08 for q ≡ 1 mod 4, 825 • 2 −16 ≈ 0.012 59 for q ≡ 3 mod 4.
As we show below, the set Σ consists of (q 2 − 8q + 15)/4 elements.Hence, a random choice of (a, b) ∈ Σ yields a maximally nonassociative quasigroup with probability ≈ 1/8.596 if q ≡ 1 mod 4, and with probability ≈ 1/19.86 if q ≡ 3 mod 4.This may have an important consequence for the cryptographic application described in [10].It means that a maximally nonassociative quasigroup of a particular large order can be obtained in an acceptable time by randomly generating pairs (a, b) until one is found for which Q a,b is maximally nonassociative.
An important ingredient in the proof of Theorem 1.1 is the transformation described in Proposition 1.2, and used in Corollary 1.3 to determine |Σ|.
Define S = S(F) as the set of all (x, y) ∈ F × F such that both x and y are squares, x y, and {0, 1} ∩ {x, y} = ∅.(1-3) [3] Maximally nonassociative quasigroups 3 The mapping The definition of Q a,b follows the established way of defining a quasigroup by means of an orthomorphism, say ψ, of an abelian group (G, +).Here, ψ is said to be an orthomorphism of (G, +) if it permutes G and the mapping The definition in Equation (1-2) of the quasigroup Q a,b thus fits the general scheme that u * v = u + ψ(v − u) is a quasigroup whenever ψ is an orthomorphism of an abelian group (G, +).See [7,15] for more information on quasigroups defined by means of orthomorphisms.
The number of associative triples in such a quasigroup depends upon the number of solutions to the associativity equation: (1-6) A. Drápal and I. M. Wanless [4] Below we always assume that ψ = ψ a,b for some (a, b) ∈ Σ.Some of our statements remain true in the case of a general ψ.However, the general situation is not the focus of this paper.
PROOF.This is a restatement of [6,Lemmas 1.3 and 3.1].A sketch of the proof follows, to make this paper self-contained.Since x is an automorphism of Q for each c ∈ F, c 0, the associativity equation holds for (u, v) if and only if it holds for (c 2 u, c 2 v).For the rest, it suffices to observe that in an idempotent quasigroup, u the respective value of i, j, r, or s is set to 1.For each (u, v) ∈ E(a, b), there hence exists exactly one quadruple (i, j, r, s) such that (u, v) ∈ E rs ij (a, b), giving us the desired partition.We also work with sets ) ∅}, where i, j, r, s ∈ {0, 1}.The next observation directly follows from the definition of the sets Σ rs ij .It is recorded here for the sake of later reference.PROPOSITION 1.5.Suppose that (a, b) ∈ Σ = Σ(F q ) for an odd prime power q > 1.The quasigroup Q a,b is maximally nonassociative if and only if (a, b) Σ rs ij . [5] Maximally nonassociative quasigroups 5 If it is assumed that (u, v) ∈ E rs ij (a, b), then the associativity equation (1-6) can be turned into a linear equation in unknowns u and v since each occurrence of ψ can be interpreted by means of Equation (1)(2)(3)(4)(5).The list of these linear equations can be found in [6].Their derivation is relatively short and is partly repeated in Lemmas 2.4-2.7.The approach used here differs from that of [6] in two aspects.The symmetries induced by opposite quasigroups and by automorphisms Q a,b Q b,a are used more extensively here, and characterizations of Σ rs ij are immediately transformed into characterizations of As will turn out, sets S rs ij can be described by a requirement that several polynomials in x and y are either squares or nonsquares.Estimates of |S rs ij | can be thus obtained by means of the Weil bound (as formulated, say, in [8,Theorem 6.22]).We are not using the Weil bound directly, but via Theorem 1.6 below, a straightforward consequence from [6,Theorem 1.4].Applications of Theorem 1.6 to the intersections of sets S rs ij , with symmetries taken into account, yield, after a number of computations, the asymptotic results stated in Theorem 1.1.
Say that a list of polynomials p 1 , . . ., p k in one variable, with coefficients in F, is square-free if there exists no sequence 1 i 1 < • • • < i r k such that r 1 and p i 1 • • • p i r is a square (as a polynomial with coefficients in the algebraic closure F of F).Define χ : F → {±1, 0} to be the quadratic character extended by χ(0) = 0. THEOREM 1.6.Let p 1 , . . ., p k ∈ F[x] be a square-free list of polynomials of degree d i 1, and let ε 1 , . . ., ε k ∈ {−1, 1}.Denote by N the number of all α ∈ F such that The purpose of Section 2 is to describe each of the sets S rs ij by a list of polynomials p(x, y) such that the presence of (x, y) ∈ S in S rs ij depends upon p(x, y) being a square or nonsquare.Theorem 2.10 gives such a description for q = |F| ≡ 1 mod 4, and Theorem 2.11 for q ≡ 3 mod 4. Section 3 contains auxiliary results that make applications of Theorem 1.6 possible.Note that Theorem 1.6 is concerned with polynomials in only one variable.To use it, one of the variables, say y, has to be fixed.If y = c, and p 1 (x, y), . . ., p k (x, y) are the polynomials occurring in Theorems 2.10 and 2.11, then Theorem 1.6 may be used without further specifications only for those c for which p 1 (x, c), . . ., p k (x, c) is a square-free list.The purpose of Section 3 is to show that this is true for nearly all c, and that the number of possible exceptional values of c is very small.Section 4 provides the estimate of S \ S rs ij for q ≡ 3 mod 4, and Section 5 for q ≡ 1 mod 4, in Theorems 4.4 and 5.5, respectively.Section 6 consists of concluding remarks.
A. Drápal and I. M. Wanless [6] 2. Quadratic residues and the associativity equation The following facts are well known [7,16] and easy to verify.LEMMA 2.1.
An alternative way to express that q ≡ 1 mod 4 is to say that −1 is a square.If * denotes the operation of the opposite quasigroup, then Working out these connections with respect to being square or nonsquare yields the following statement.It appears without a proof since it coincides with [6, Lemmas 3.2 and 3.3] and since the proof is straightforward.LEMMA 2.2.Assume (a, b) ∈ Σ and i, j, r, s ∈ {0, 1}.Then (2-2) Both of the mappings (x, y) → (y, x) and (x, y) → (x −1 , y −1 ) permute the set S = S(F).If i, j, r, s ∈ {0, 1}, then By definition, (x, y) ∈ S if and only if x and y are both squares, x y, and {x, y} ∩ {0, 1} = ∅.These properties are retained both by the switch (x, y) → ( y, x) and by the inversion (x, y) → (x −1 , y −1 ).These mappings thus permute S.
For the proof, we thus need to show that Maximally nonassociative quasigroups 7 To determine all of the sets S rs ij , it thus suffices to know the sets S 00 00 , S 01 00 , S 11 00 , S 00 01 , S 01 01 , and S 10 01 . (2-4) We next determine these sets via a sequence of lemmas.PROOF.We assume that −1 is a square.If (u, v) ∈ E 00 00 (a, b), then the associativity equation attains the form a(au − v) = −av + a(u − v + av), and that is the same as (1−a)(u−v) = 0. Since 1−a 0, and since u is assumed to be square, the set E 00 00 (a, b) is nonempty if and only if it contains (1, 1), by Proposition 1.4.This takes place if and only if 1−a and a are squares.Suppose that is always a square, the conditions for the existence of the solution are that a − v and 1 This implies that uv is a square.However, the assumption (u, v) ∈ E 00 01 (a, b) implies that u is a square and −v is a nonsquare.Thus, uv should be both a square and a nonsquare, which is a contradiction.PROOF.We assume that −1 is a nonsquare.If E 00 00 (a, b) ∅, then (1, 1) ∈ E 00 00 (a, b), by the same argument as in the proof of Lemma 2.4.However, (1, 1) cannot belong to a − a/b and 1 . Thus, (1−y)(x−y) has to be a nonsquare.

PROOF. In this case, the associativity equation yields a(au
01 (a, b), and one of the elements a 2 − b and b 2 − 2b + a is equal to zero, then the other has to vanish as well.Assume that Computations above show that [9] Maximally nonassociative quasigroups 9 Suppose now that at least one of , and .
This is true for each prime power q 47.REMARK 2.8.Lemmas 2.4-2.7 cover all sets S rs ij that are listed in (2)(3)(4).Up to the exceptions discussed in Remark 2.9, each of these sets is either empty, or is described by a list of polynomials, say p 1 , . . ., p k ∈ F[x, y], k ∈ {2, 3}, and elements ε h ∈ {−1, 1}, such that (ξ, η) ∈ S belongs to S rs ij if and only if χ(p h (ξ, η)) = ε h , for 1 h k.This is because the polynomials p h (x, y) have been determined in all cases in such a way that if would be equal to zero, and that is impossible, by Proposition 1.4.
Note that (ξ, η) was used in Remark 2.8 to emphasize the distinction between elements of S and formal variables x and y.In the remainder of the paper, elements of S will again be denoted by (x, y).The context will always be clear.REMARK 2.9.Sets S 01 01 and S 10 10 behave exceptionally in the sense that the regular behavior described in Remark 2.8 needs an assumption that y+1−x 0 or x 2 −x−1 0 (for the set S 01 01 ), and that x+1−y 0 or y 2 −y−1 0 (for the set S 10  10 ).There are at most A. Drápal and I. M. Wanless [10] two pairs (x, y) ∈ S such that y+1−x = 0 = x 2 −x−1 and at most two pairs (x, y) ∈ S such that x+1−y = 0 = y 2 −y−1.Hence, assuming that causes no difficulty when estimating σ(q).If Condition (2-5) does not hold, then (x, y) ∈ S 01 01 ∪ S 10 10 if q 47, by point (i) of Lemma 2.7.In fact, if Condition (2-5) does not hold, then (x, y) ∈ S rs ij for each q 3, by [6] (compare with the application of [6, Lemma 3.4] in the proof of [6, Theorem 3.5]).
For p(x, y) ∈ F[x, y] such that x p(x, y) and y p(x, y), define the reciprocal polynomial p(x, y) as x n y m p(x −1 , y −1 ), where n and m are the degrees of the polynomial p in the variables x and y, respectively.Note that if (x, y) ∈ S, then χ( p(x, y)) = χ(x n y m p(x −1 , y −1 )) = χ(p(x −1 , y −1 )) since x and y are squares.Note also that p(x, y) = p(x, y), 1 A description of those sets S rs ij that do not occur in List (2-4) can be derived from Lemmas 2.4-2.7 by means of Proposition 2. Following this pattern, a characterization of all sets S rs ij may be derived from Lemmas 2.4-2.7 by means of Proposition 2.3.This is done in Theorems 2.10 and 2.11.Since the derivation is straightforward, both of them are stated without a proof.Set Note that g 3 (x, y) = ĝ1 (x, y), g 4 (x, y) = ĝ2 (x, y) = g 3 ( y, x), and g 2 (x, y) = g 1 ( y, x).THEOREM 2.10.Assume that q ≡ 1 mod 4 is a prime power, and that S = S(F q ).Let (x, y) ∈ S be such that Condition (2)(3)(4)(5)   THEOREM 2.11.Assume that q ≡ 3 mod 4 is a prime power, and that S = S(F q ).Let (x, y) ∈ S be such that Condition (2)(3)(4)(5)

Avoiding squares
Our goal is to estimate the size of the set T = S \ S rs ij .Since Theorem 1.6 requires polynomials in one variable, to determine the size of T, it is necessary to proceed by determining the sizes of slices {x ∈ F : (x, c) ∈ T} for each square c {0, 1}.As a convention, p(x, c) will mean a polynomial in one variable, that is, an element of F[x] for every p(x, y) ∈ F[x, y].
Theorem 1.6 may be directly applied only when the product of the polynomials involved is square-free.Thus, for p 1 (x, y), . . ., p k (x, y) ∈ F[x, y], it is necessary to set aside those c ∈ F for which p 1 (x, c), . . ., p k (x, c) is not a square-free list of polynomials.An asymptotic estimate does not depend upon the number of c set aside if there is only a bounded number of them.Hence, a possible route is to express the discriminant of p 1 (x, c) • • • p k (x, c) by means of computer algebra, and then set aside those c that make the discriminant equal to zero.The route taken below is elementary and is not dependent upon a computer.In this way, the number of c to avoid is limited to 51. [12] This is a consequence of the following statement, the proof of which is the goal of this section.THEOREM 3.1.Let F be a field of characteristic different from 2. The list of polynomials is square-free if the following conditions hold: c is a root of neither x 2 −3x+3 nor 3x 2 −3x+1; c is a root of neither x 3 −2x 2 +3x−1 nor x 3 −3x 2 +2x−1. (3-11) The proof requires a number of steps.As an auxiliary notion, we call a list of polynomials p 1 (x, y), . . ., p k (x, y) ∈ F[x, y] reciprocally closed if for each i ∈ {1, . . ., k}, both x p i (x, y) and y p i (x, y) are true, and there exist unique j ∈ {1, . . ., k} and λ ∈ F such that pi (x, y) = λp j (x, y).
If a = a i t i ∈ F[t] is a nonzero polynomial of degree d 0, then the reciprocal polynomial a i t d−i will be denoted by â, like in the case of two variables.A list a 1 (t), . . ., a k (t) ∈ F[t] is reciprocally closed if for each i ∈ {1, . . ., k}, the polynomial a i (t) is not divisible by t, and there exist unique j ∈ {1, . . ., k} and λ ∈ F such that âi (t) = λa j (t).LEMMA 3.2.Let p 1 (x, y), . . ., p k (x, y) ∈ F[x, y] and a 1 (t), . . ., a r (t) ∈ F[t] be two reciprocally closed lists of polynomials.Denote by Γ the set of all nonzero roots of polynomials a 1 , . . ., a r .Assume that holds for all i ∈ {1, . . ., k}.Let i, j ∈ {1, . . ., k} and λ ∈ F be such that p j (x, y) = λ pi (x, y) and i j.If gcd(p i (x, c), p (x, c)) = 1 https://doi.org/10.1017/S1446788722000386Published online by Cambridge University Press [13] Maximally nonassociative quasigroups 13 holds for all nonzero c ∈ F \ Γ and all i, 1 k, then holds for all nonzero c ∈ F \ Γ and h j, 1 h k.
PROOF.Suppose that h j and c ∈ F \ Γ, c 0, are such that p j (x, c) and p h (x, c) have a common root in F, say γ.Thus, p j (γ, c) = p h (γ, c) = 0.By Condition (3-12), γ 0. Since the list p 1 (x, y), . . ., p k (x, y) is reciprocally closed, there exists i such that p (x, y) is a scalar multiple of ph (x, c).Since p j (x, y) is a multiple of pi (x, y), we have p i (γ −1 , c −1 ) = 0 = p (γ −1 , c −1 ) and hence gcd(p i (x, c −1 ), p (x, c −1 )) 1.By the assumption on p i , this cannot be true unless c −1 ∈ Γ.We refute the latter possibility by proving that if c −1 ∈ Γ, then c ∈ Γ.That follows straightforwardly from the assumption that the list a 1 , . . ., a r is reciprocally closed.Indeed, since c −1 ∈ Γ, there exists s ∈ {1, . . ., r} such that a s (c −1 ) = 0.There also exists m ∈ {1, . . ., r} such that a m is a scalar multiple of âs .Because of that, a m (c) = a m ((c −1 ) −1 ) = 0.This implies that c ∈ Γ since Γ is defined as the set of all nonzero roots of polynomials a 1 , . . ., a r .
If a(t) = t − γ, γ 0, then â(t) = −γ(t − γ −1 ).Hence, the list of nonzero c that fulfill one of the conditions (3-2)-(3-11) may be considered as a set Γ of nonzero roots of a reciprocally closed list of polynomials in one variable.Now, remove x and x−1 from the list of polynomials (3-1) that are the input to Lemma 3.2.The remaining polynomials can be interpreted as a list p 1 (x, c), . . ., p 13 (x, c) such that p 1 (x, y), . . ., p 13 (x, y) is a reciprocally closed list of polynomials in two variables.It is easy to verify that if 0 or 1 is a root of any of the polynomials p i (x, c), 1 i 13, then c fulfills Condition (3-2).Polynomials x and x−1 can be thus excised from the subsequent discussion, and Lemma 3.2 may be used.Lemma 3.2 will also be applied to some sublists of p 1 (x, c), . . ., p 13 (x, c) that are reciprocally closed.The first such sublist is the linear polynomials occurring in (3-1) (with x and x−1 being removed).These are , and (1−2c)x + c 2 .The latter two polynomials are equal to g 2 (x, c) and g 4 (x, c).The list of these linear polynomials is square-free if there are no duplicates in the set of their roots The reciprocity yields the following pairs of roots: We now prove a sequence of lemmas which explore properties of the polynomials in List (3-1).PROOF.By Lemma 3.2, it suffices to consider only the polynomial h(x) = g 1 (x, c).(3)(4)(5), and (3-7)- (3)(4)(5)(6)(7)(8)(9)(10)(11), then none of the elements of R(c) is a root of f i (x, c) for any i = 1, 2, 3, 4.
PROOF.The proof is very similar to that of Lemma 3.5, so we only give a summary.By Lemma 3.2, it suffices to test the polynomials f 1 (x, c) and f 2 (x, c).Substituting an element of R(c) in place of x always yields a polynomial from the indicated list.Note that , and . [15] Maximally nonassociative quasigroups 15 LEMMA 3.7.Suppose that c ∈ F satisfies Conditions  and (3)(4)(5).Then for each i ∈ {1, 3}, there exist at least three j ∈ {1, 2, 3, 4} such that g i (x, c) and f j (x, c) share no root in F.
PROOF.Because of the reciprocity, i = 1 may be assumed.If PROOF.This is obvious if c).This is why c 2 −1 has to be assumed.
For the rest, it suffices to test pairs (1, 2) and (2, 3), by the reciprocity described in Lemma 3.2.
We can now bring all the pieces together to prove the main result of this section.

When −1 is a nonsquare
Throughout this section, F = F q will be a finite field of order q ≡ 3 mod 4. We put Σ = Σ(F q ) and S = S(F q ).By Corollary 1.3, |S| = |Σ| = (q 2 −8q+15)/4.Define Srs ij = S \ S rs ij and T = Srs ij , where sets S rs ij are characterized by Theorem 2.11, subject to the assumption that Condition (2-5) holds.As we can see, Condition (2-5) holds in all cases that are relevant for our calculations.The aim of this section is to estimate the number σ(q) = |{(a, b) ∈ Σ : Q a,b is maximally nonassociative}|.By Proposition 1.5, A. Drápal and I. M. Wanless [16] σ and define T 0 , T 1,1 , T 1,−1 , and T 2 by exchanging x and y.For example, this means that All of these unions are unions of disjoints sets.Both of the mappings (x, y) → ( y, x) and (x, y) → (x −1 , y −1 ) permute T. Both of them exchange T 1 and T 1 , and T 2 and T 2 .Furthermore, (x, y) PROOF.Recall that by our definition of S, we have 1 {x, y} and x y for all (x, y) ∈ T. By Proposition 2.3, both (x, y) → ( y, x) and (x, y) → (x −1 , y −1 ) permute T. The effects of these two mappings are simple to verify.Note, for example, that if ε To see that In the next two propositions, we seek estimates of these quantities.In both results, we assume that c fulfills Condition .Observe that under this assumption, c 2 −c−1 0 and for all x ∈ F q , either x c+1 or x 2 −x−1 0, and therefore Condition (2-5) holds for (x, y) = (x, c).This will enable us to use Theorem 2.11.PROPOSITION 4.2.Suppose that c and 1 − c are both nonzero squares in F q and that c fulfills Conditions (3-2)- (3)(4)(5)(6)(7)(8)(9)(10)(11).Then, PROOF.We estimate t 2 (c) by characterizing the pairs (x, c) in T 2 .For a fixed c, there are at most 21 values of x that are roots of any of the polynomials in List (3-1).So at the cost of adding a term equal to 21 to our eventual bound, we may assume for the remainder of the proof that x is not a root of any polynomial in List (3-1).Then From the definitions of S00 01 , S00 10 , S11 10 , and S11 01 , we deduce that χ(g 1 (x, c)) = χ(g 4 (x, c)) = −1 and χ(g 2 (x, c)) = χ(g 3 (x, c)) = 1.Now, from (x, c) ∈ S01 01 , we deduce that either (4-1) [19] Maximally nonassociative quasigroups 19 5. When −1 is a square Throughout this section, F = F q will be a finite field of order q ≡ 1 mod 4. Our broad strategy for obtaining an estimate of σ(q) is similar to that used in Section 4. For i, j, r, s ∈ {0, 1}, define Srs ij = S \ S rs ij and put T = Srs ij .The set T will again be expressed as a disjoint union of sets, the size of each of which can be estimated by means of the Weil bound.Let ε = χ(x − y) and define We write R( ρ) as a shorthand for R( ρ 1 , ρ 2 , ρ 3 , ρ 4 ), where ρ = ( ρ 1 , ρ 2 , ρ 3 , ρ 4 ).We record the following basic facts about the sets just defined.LEMMA 5.1.Suppose ρ j ∈ {−1, 1} for 1 j 4. The map (x, y) → ( y, x) induces bijections that show that PROOF.By Proposition 2.3, we know that (x, y) → (x −1 , y −1 ) permutes each of the sets T 1 , T 2 , and T 2 , while (x, y) → ( y, x) permutes T 1 and swaps T 2 and T 2 .This gives us a bijection between T 2 and T 2 .Note also that The remaining claims about bijections follow directly from the definitions of f 1 , f 2 , f 3 , and f 4 in Equations (2)(3)(4)(5)(6).
If (x, y) ∈ R( ρ 1 , ρ 2 , −1, −1), then χ( f j (x, y)) = −ε for both j ∈ {3, 4}.That implies (x, y) ∈ S 11 00 , by Lemma 2.4.Hence, R( ρ 1 , ρ 2 , −1, −1) = ∅ and our bijection gives Our aim is to use the |R i ( ρ)| to estimate the size of T. We should note that T may be a proper superset of ρ R( ρ).The (small) difference arises from the contribution to T from roots of the polynomials f i (this contribution will be accounted for later, when all roots are included as an error term in our bounds).Lemma 5.1 reduces the number of |R i ( ρ)| that we need to estimate to only those ρ shown in Table 1.The final column of that table shows the multiplicity μ that we need to use for each Our observations above allow us to symbolically describe polynomial lists for each of the sets R 1 ( ρ).We have Combining this information with the last column of Table 1, we reach a symbolic description of the polynomials contributing to t 1 (c) that contains (10,15) with multiplicity 2, (11,15) with multiplicity 1 + 4 + 2 = 7, (13, 18) with multiplicity 4 • 3 + 2 • 6 + 2 • 6 = 36, and (15, 21) with multiplicity 2 • 9 = 18.In each case, the list of polynomials involved is square-free, by Theorem 3.1.Hence, we may apply Theorem 1.6 to find that   We are now ready to prove the main result for this section.
THEOREM 5.5.If q ≡ 1 mod 4, then PROOF.There are (q − 3)/2 choices for a square c ∈ F q satisfying c {0, 1}.At most, 49 of these choices do not fulfill Conditions Next, by Lemma 5.1, we know that The result then follows from simple rearrangement.
COROLLARY 5.6.Let q run through all prime powers that are congruent to 1 mod 4.

Concluding remarks
Theorems 4.4 and 5.5 give formulas that can be used as estimates of σ(q) for large q.We did not work hard to optimize the constants in the bounds.Even if we had, the number of applications of the Weil bound is too big to allow the estimates to be useful for small q.However, for large q, our results show that maximally nonassociative quasigroups can be generated via random sampling.If (a, b) is chosen uniformly at random from Σ(F q ), then it can quickly be checked (using O(1) evaluations of χ, as shown in Theorems 2.10 and 2.11) whether Q a,b is maximally nonassociative, and the A. Drápal and I. M. Wanless [24] probability of success is bounded away from zero.We thus have a computationally realistic method of generating random maximally nonassociative quasigroups of large orders.This might be of interest, given the cryptographic applications [10].
The approach that we have used in this paper might be adapted to resolve [3, Conjecture 5.10], which is concerned with the density of parameters that yield a maximally nonassociative quasigroup when constructing such a quasigroup by means of a nearfield.
In a future paper, we plan to consider how many different isomorphism classes are represented by the maximally nonassociative quasigroups generated from quadratic orthomorphisms.To answer this question requires theory to be developed on when different quadratic orthomorphisms generate isomorphic quasigroups (which is a question of independent interest).Some limited circumstances where different orthomorphisms create isomorphic quasigroups are already known [16].In particular, we know that Q a,b Q b,a (see Lemma 2.1(i)), and that Q a,b Q a p ,b p in any field of characteristic p.We expect these to generate the only isomorphisms that affect the asymptotics.In other words, we conjecture that the number of quasigroups (up to isomorphism) is asymptotic to σ(q)/(2 log p q), where σ(q) is estimated by Theorems 4.4 and 5.5.
In our analysis leading to our main results, we discarded all roots of the polynomials in List (3-1).Bajtoš [1] has investigated these cases by finding the asymptotic number of solutions to t(x, y) = 0 when t is one of the polynomials x−1−y, x−xy−y, y−1−x, y−xy−x, f j (x, y), or g j (x, y) for 1 j 4. For each of these polynomials, Bajtoš determines the density of parameters (a, b) that yield a maximally nonassociative quasigroup.The density is measured with respect to the size of the set of all (a, b) for which the given polynomial gives zero, with x = a/b and y = (1−a)/(1−b).For each polynomial, that set has size asymptotically equal to q/4.Because of symmetry and reciprocity of polynomials, it suffices to investigate cases (a) x−1−y, (b) x 2 +y 2 −xy−x, and (c) x 2 +y−2x.In case (a), the obtained densities are ≈ 0.109, 0.219, 0.031, and 0.047, with q ≡ 1, 5, 3, 7 mod 8, respectively.For case (b), the densities are 0.109 and 0.082, q ≡ 1, 3 mod 4. Case (c) yields 0.156, 0.172, 0.047, and 0.031, for q ≡ 1, 5, 3, 7 mod 8.These numbers thus give probabilities of finding a maximally nonassociative quasigroup by a random choice, for each of the investigated cases.In the general case, these probabilities are ≈ 0.116 and 0.050 for q ≡ 1, 3 mod 4 (compare with the comments following Theorem 1.1).
Maximally non-associative quasigroups minimize the number of associative triples.Gowers and Long [9] consider another measure of how associative a quasigroup is, which they call its number of 'octahedra.' Several interesting connections between the number of octahedra and the number of associative triples are shown in their work, although they concentrate on quasigroups which are in some sense close to associative.Subsequently, Kwan et al. [12] considered the typical number of octahedra in a random quasigroup and asked a question regarding how few octahedra a quasigroup of order n can have.The maximally nonassociative quasigroups that we have constructed may be useful in answering that question, but the connection requires further investigation.Another open question is how few nonassociative triples loops (quasigroups with identity) can have.It is not difficult to show that a loop of order n has to possess at least 3n 2 − 3n + 1 associative triples.However, presently, no loop with less than 3n 2 − 2n triples seems to be known.In [2], quadratic orthomorphisms were used to construct loops of order n = p + 1, for a prime p 13, that have exactly 3n 2 − 2n associative triples.The chosen method failed for p = 19.That case was solved by means of a ternary orthomorphism, which leads us naturally to our last research direction.
The question of when maximally nonassociative quasigroups can be generated by orthomorphisms that are not quadratic is wide open.Some examples are given in [2,6].Perhaps the next case to study would be orthomorphisms that are cyclotomic but not quadratic.We finish with some examples of this type that produce maximally nonassociative quasigroups.Each orthomorphism is given as a permutation in cycle notation.We have (1,3,8)

For
(a, b) ∈ Σ, denote by E(a, b) the set of (u, v) (0, 0) that satisfy the associativity equation (1-6).By Proposition 1.4, Q a,b is maximally nonassociative if and only if E(a, b) = ∅.The number of such (a, b) may be obtained indirectly by counting the number of (a, b) ∈ Σ for which E(a, b) ∅.To this end, we partition E(a, b) = E rs ij (a, b), where i, j, r, s ∈ {0, 1}.To determine to which part an element (u, v) ∈ E(a, b) belongs, the following rule is used: