Sums of two squares are strongly biased towards quadratic residues

Chebyshev famously observed empirically that more often than not, there are more primes of the form $3 \bmod 4$ up to $x$ than of the form $1 \bmod 4$. This was confirmed theoretically much later by Rubinstein and Sarnak in a logarithmic density sense. Our understanding of this is conditional on the generalized Riemann hypothesis as well as on the linear independence of the zeros of $L$-functions. We investigate similar questions for sums of two squares in arithmetic progressions. We find a significantly stronger bias than in primes, which happens for almost all integers in a \emph{natural density} sense. Because the bias is more pronounced, we do not need to assume linear independence of zeros, only a Chowla-type conjecture on nonvanishing of $L$-functions at $1/2$. To illustrate, we have under GRH that the number of sums of two squares up to $x$ that are $1 \bmod 3$ is greater than those that are $2 \bmod 3$ 100% of the time in natural density sense.


Introduction 1A Review of sums of two squares in arithmetic progressions
Let S be the set of positive integers expressible as a sum of two perfect squares.We denote by 1 S the indicator function of S. It is multiplicative and for a prime p we have 1 S (p k ) = 0 if and only if p ≡ 3 mod 4 and 2 k. (1.1) Landau [Lan08] proved that where K = p≡3 mod 4 (1−p −2 ) −1/2 / √ 2 ≈ 0.764 is the Landau-Ramanujan constant.See [Har40, Lec.IV] for Hardy's account of Ramanujan's unpublished work on this problem.Landau's method yields an asymptotic expansion in descending powers of log x, which gives an error term O k (x/(log x) k+1/2 ) for each k ≥ 1. 1  Prachar [Pra53] proved that sums of two squares are equidistributed in arithmetic progressions, in the following sense.If (a, q) = 1 then S(x; q, a) := #{n ∈ S : n ≤ x, n ≡ a mod q} ∼ (4, q) (2, q)q p|q p≡3 mod 4 as x → ∞ as long as a ≡ 1 mod (4, q); see Iwaniec's work on the half-dimensional sieve for results allowing q to vary with x [Iwa76].The condition a ≡ 1 mod (4, q) is necessary: otherwise (a, q) = 1 and a ≡ 1 mod (4, q) imply a ≡ 3 mod 4.However, S is disjoint from 3 mod 4.

1B Main theorem and corollary
Here we consider a Chebyshev's bias phenomenon for S. We ask, what can be said about the size of the set {n ≤ x : S(n; q, a) > S(n; q, b)} (1.2) for distinct a, b mod q with a ≡ b ≡ 1 mod (4, q) and (a, q) = (b, q) = 1?These conditions guarantee that S(n; q, a) ∼ S(n; q, b) → ∞ as n → ∞, so it is sensible to study (1.2).We let χ −4 be the unique nonprincipal Dirichlet character modulo 4. Motivated by numerical evidence (based on n ≤ 10 8 ) showing S(n; 3, 1) − S(n; 3, 2) and S(n; 5, 1) − S(n; 5, 3) are positive much more frequently than not, we were led to discover and prove the following.
If χ(a) = 1 for all real characters modulo q then a is a quadratic residue modulo q, and vice versa.Let us specialize a to be a quadratic residue modulo q and b to be a nonquadratic residue.We observe that (on GRH) (χ(a) − χ(b)) L(1/2, χ)L(1/2, χχ −4 ) is nonnegative for each real χ mod q that is not χ 0 or χ 0 χ −4 .Hence, under GRH and our assumptions on a and b, a necessary and sufficient condition for (1.3) to hold is that L(1/2, χ)L(1/2, χχ −4 ) = 0 for some real χ mod q with χ(b) = −1.One way to guarantee this is to assume Chowla's Conjecture.We state this as the following corollary.
It would be interesting to try and establish the positivity of (1.3), possibly in a statistical sense, without hypotheses like Chowla's Conjecture.
For a given Dirichlet character χ, one can computationally verify that L(1/2, χ) is nonzero, and in fact compute all zeros of L(s, χ) up to a certain height, see Rumely [Rum93] which in particular shows L(1/2, χ) = 0 for characters of conductor ≤ 72.Nowadays computing L(1/2, χ) is a one-line command in Mathematica, and so the verification of (1.3) is practical for fixed q, a and b.
We expect the expression in (1.3) to be nonzero as long as χ(a) = χ(b) for some real character, or equivalently, if a/b is nonquadratic residue modulo q.It is instructive to consider the following two possibilities for a and b separately: • Suppose a, b and a/b are all nonquadratic residues, a situation that could occur only if the modulus q is composite.Although the expression C q,a,b should be nonzero and give rise to a bias, it seems the sign is very difficult to predict.Interestingly, for primes, as we shall review below, there is no bias in this case.
• If exactly one of a and b is a quadratic residue then Corollary 1.3 tells us the direction of the bias (if it exists) is towards the quadratic residue.A sufficient condition for the bias to exist is Chowla's Conjecture.

1C Comparison with primes
Chebyshev's bias was originally studied in the case of primes, that is, replacing S by the set of primes.Letting π(x; q, a) be the numbers of primes up to x lying in the arithmetic progression a mod q, Chebyshev famously observed in 1853 that π(x; 4, 3) > π(x; 4, 1) happens more often than not [Tch62,.
Rubinstein and Sarnak [RS94] studied the set {x : π(x; q, a) > π(x; q, b)} where a ≡ b mod q and (a, q) = (b, q) = 1.They showed, under GRH and the Grand Simplicity Hypothesis (GSH) that this set has logarithmic density strictly between 0 and 1.Additionally, the logarithmic density is greater than 1/2 if and only if a is a nonquadratic residue and b is a quadratic residue.In particular, no bias is present at all if both a and b are nonquadratic residues, as opposed to the sums of two squares analogue.
GSH asserts that the multiset of γ ≥ 0 such that L(1/2 + iγ, χ) = 0, for χ running over primitive Dirichlet characters, is linearly independent over Q; here γ are counted with multiplicity.It implies Chowla's Conjecture (since 0 is linearly dependent) and that zeros of L(s, χ) are simple.As opposed to Chowla, it is very hard to gather evidence for GSH, even for individual L-functions.However, see Best and Trudgian for such evidence in the case of ζ [BT15].In the literature, this hypothesis also goes under the name Linear independence (LI).
As far as we are aware, Theorem 1.1 is the first instance where a set of integers of arithmetic interestin this case sums of two squares -is shown to exhibit a complete Chebyshev's bias, that is, a bias that holds for a natural density-1 set of integers: where M (n; q, a) counts elements up to n in a set M ⊆ N that are congruent to a modulo q.A key issue here is the natural density: Meng [Men20] has a related work about a bias that holds for a logarithmic density-1 set of integers (see §2E).Recently, Devin proposed a conjecture [Dev21, Conj.1.2] on a bias in logarithmic density 1. See Fiorilli [Fio14b,Fio14a] for biases, in logarithmic density, that come arbitrarily close to 1, and Fiorilli and Jouve [FJ22] for complete biases in 'Frobenius sets' of primes (that generalize arithmetic progressions).
We also mention a very strong bias was proved by Dummit, Granville and Kisilevsky [DGK16] who take Chebyshev's observation to a different direction.They show that substantially more than a quarter of the odd integers of the form pq up to x, with p, q both prime, satisfy p ≡ q ≡ 3 mod 4.
2 Origin of the bias, computational evidence and a variation 2A Review of the original bias Fix a modulus q.All the constants below might depend on q.We give an informal explanation for the origin of the bias.It is instructive to start with the case of primes.By orthogonality of characters, π(x; q, a) − π(x; q, b) = 1 φ(q) The generating function of primes was studied by Riemann [Rie59], who showed that for s > 1.Here µ is the Möbius function, ζ is the Riemann zeta function and the logarithm is chosen so that log ζ(s) is real if s is real and greater than 1.More generally, given a Dirichlet character χ we have We may also write this L-function identity in terms of arithmetic functions: where Λ is the von Mangoldt function and α is supported only on cubes and higher powers and its sum is negligible for all practical purposes.Under GRH we can show that (see [RS94, Lem.2.1]) and (still under GRH) we can use the explicit formula to show that n≤x χ(n)Λ(n) is typically of order √ x, in the sense that 1 see [MV07,Thm. 13.5].Under linear independence, one can show that the random variable has a limiting distribution with expected value 0 (here y is chosen uniformly at random between 0 and Y , and Y → ∞).The exponential change of variables leads to the appearance of logarithmic density.To summarize, (log x/ √ x) n≤x χ(n)Λ(n)/ log n (with x = e y ) has expectation 0 and order of magnitude 1.The bias comes from the term −1 n=p 2 , p prime χ 2 (p)/2.Indeed, If χ is a nonreal character, χ 2 is nonprincipal and GRH guarantees this sum is o( √ x/ log x).However, if χ is real, this sum is of the same order of magnitude as n≤x χ(n)Λ(n)/ log n, namely it is using the Prime Number Theorem.
Rubinstein and Sarnak replaced n≤x χ(n)Λ(n)/ log n with n≤x χ(n)Λ(n)/ log x using partial summation.This is advantageous as we have a nice explicit formula for the sum of χ(n)Λ(n).However, one can work directly with χ(n)Λ(n)/ log n, whose generating function is log L(s, χ), and this was done by Meng [Men18] in his work on Chebyshev's bias for products of k primes.Meng's approach is more flexible because it works even when the generating function has singularities which are not poles.So while −L (s, χ)/L(s, χ) = n≥1 Λ(n)χ(n)/n s is meromorphic with simple poles at zeros of L(s, χ), which leads to the explicit formula by using the Residue theorem, Meng's approach can deal with − log L(s, χ) directly although it does not have poles, rather it has essential singularities.Meng's work applies in particular to k = 1 and k = 2, thus generalizing Rubinstein and Sarnak as well as Ford and Sneed [FS10].

2B The generating function of sums of two squares
Let us now return to sums of two squares.Let a, b be residues modulo q with (a, q) = (b, q) = 1 and a ≡ b ≡ 1 mod (4, q).Orthogonality of characters shows (2.2) We want to relate n≤x 1 S (n)χ(n) to L-functions and their zeros, and obtain an analogue of the explicit formula for primes.The generating function of sums of two squares was studied by Landau [Lan08], who showed that for s > 1, where H has analytic continuation to s > 1/2.Here the square root is chosen so that ζ(s) and L(s, χ −4 ) are real and positive for s real and greater than 1.This representation of the generating function plays a crucial role in the study of the distribution of sums of two squares, see e.g.[GR21].Later, Shanks [Sha64, p. 78] and Flajolet and Vardi [FV96, pp.7-9] (compare [Rad14, Eq. (3)], [GR21, Lem.2.2]) proved independently the identity and their proof can yield an analogue of (2.4) with a twist by χ(n).Shanks and Flajolet and Vardi were interested in efficient computation of the constant K, and this identity leads to Since both sides of (2.4) enjoy Euler products, this identity can be verified by checking it locally at each prime; one needs to check p = 2, p ≡ 1 mod 4 and p ≡ 3 mod 4 separately using (1.1).For the purpose of this paper we do not need the terms corresponding to k > 1 in (2.4).What we need is stated and proved in Lemma 3.5, namely that for G which is analytic and nonvanishing in s > 1/4 and bounded in s ≥ 1/4 + ε for each ε > 0. The important feature of this formula is that it allows us to analytically continue F (s, χ) to the left of s = 1/2 (once we remove certain line segments), as opposed to (2.3) whose limit is s > 1/2.See the discussion at the end §3B.
Recently, a formula very similar to (2.5) was used by Porritt [Por20] in his study of character sums over polynomials with k prime factors, and k tending to ∞.We state his formula in the integer setting.Let Ω be the additive function counting prime divisors with multiplicity.He showed that, for complex z with |z| < 2, we have [Por20, Eq. ( 4)] for E z (s, χ) which is analytic in s > max{1/3, log 2 |z|}.He then proceeds to apply a Selberg-Delange type analysis, leading to an explicit formula for a polynomial analogue of n≤x, Ω(n)=k χ(n) where k grows like a log log x for a ∈ (0, 2 1/2 ) (in the polynomial world, q and q 1/2 replace 2 and 2 1/2 , where q is the size of the underlying finite field).His results show a strong Chebyshev's bias once a > 1.2021 . . .[Por20,Thm. 4].

2C Analyzing singularities
We shall analyze each of the sums in (2.2).We first observe that we do not need to analyze the sums corresponding to χ or χχ −4 being principal, because these characters do not contribute to (2.2) (as χ 0 (a) = χ 0 (b)).
We call these zeros and poles 'singularities of F '.They all lie on s = 1/2.We construct an open set A χ by taking the half-plane s > 1/4 are removing the segments {σ + it : 1/4 < σ ≤ 1/2} for every singularity 1/2 + it.This domain is simply connected and L(s, χ), L(s, χχ −4 ), L(2s, χ 2 ) and L(2s, χ 2 χ −4 ) have no poles or zeros there.Hence, they have well-defined logarithms there and we may analytically continue F (s, χ) to A χ .Although we cannot literally shift the contour to the left of s = 1/2, we can move to a contour which stays in A χ and is to the left of s = 1/2 'most of the time'.Specifically, we shall use truncated Hankel loop contours going around the singularities, joined to each other vertically on s = 1/2 − c, as in Meng [Men18,Men20].The precise contour is described in §3D.See Figure 1 for depiction.A truncated Hankel loop contour around a singularity ρ of F (s, χ) is a contour H ρ traversing the path depicted in Figure 2. It is parametrized in (3.7).Given a character χ and a singularity ρ of F (s, χ) we let f (ρ, χ, x) := 1 2πi Hρ F (s, χ)x s ds s be the Hankel contour integral around ρ.By analyticity, the value of f (ρ, χ, x) is independent of r (once r is small enough) and for our purposes we choose r = o(1/ log x).We end up obtaining (2.6) See Lemma 3.10 for a statement formalizing (2.6) (in practice we sum only over zeros up to a certain height).If F (s, χ) ∼ C(s − ρ) m asymptotically as s → ρ + (i.e. as s − ρ tends to 0 along the positive part of the real as x → ∞.An informal way to see this is z m e z dz and the last integral gives (2.7) by Hankel's original computation [Ten15, Thm.II.0.17].2In particular, from analyzing L(s, χ)L(s, χχ −4 ) we see that we have , for any given ρ = 1/2 where m ρ is the multiplicity of ρ in L(s, χ)L(s, χχ −4 ).As m ρ ≥ 1, we are led to think of ρ =1/2 f (ρ, χ, x) as a quantity of order √ x(log x) −3/2 .However, we are not able to use (2.7) in order to bound ρ =1/2 f (ρ, χ, x) efficiently.We proceed via a different route and show without making use of (2.7).It follows that ρ =1/2 f (ρ, χ, x) √ x(log x) −3/2 most of the time.This estimate is analogous to (2.1) and its proof is similar too.We believe this sum is ε √ x(log x) ε−3/2 always but are not able to show this.This is similar to how GRH can show n≤x Λ(n)χ(n) √ x(log x) 2 , but the true size in this question is expected to be ε √ x(log x) ε .Here one should think of f (ρ, χ, x) as an analogue of the expression −x ρ /ρ from the explicit formula.

2E Martin's conjecture
Let ω be the additive function counting prime divisors (without multiplicity).In [Men20], Meng states a conjecture of Greg Martin, motivated by numerical data, saying that contains all sufficiently large x.Meng assumed GRH and GSH to prove that this set has logarithmic density 1.He also obtains results for other moduli under Chowla's Conjecture, and studies an analogous problem with the completely additive function Ω. Meng writes: 'In order to prove the full conjecture, one may need to formulate new ideas and introduce more powerful tools to bound the error terms of the summatory functions' [Men20,Rem. 4].We are able to prove a natural density version of Meng's result, making progress towards Martin's conjecture.We do not assume GSH.
Theorem 2.1.Fix a positive integer q.Assume that GRH holds for the Dirichlet L-functions L(s, χ) for all Dirichlet character χ modulo q.Then, whenever a, b satisfy (a, q) = (b, q) = 1 and D q,a,b := The proof is given in §6.If a is a quadratic residue modulo q and b is not, a sufficient condition for D q,a,b to be positive is Chowla's Conjecture.

Preparatory lemmas
Given a Dirichlet character χ modulo q we write This converges absolutely for s > 1.For s = 1 (or smaller) it does not, because such convergence implies p ≡3 mod 4, p q 1/p converges.Observe that F (s, χ) does not vanish for s > 1.We shall use the convention where σ and t denote the real and complex parts of s ∈ C.
This lemma leads to the following, which is a variation on [Ten15, Cor.II.2.4].
with an absolute implied constant.
Proof.Since |a n | ≤ 1 we have σ a ≤ 1 so that we may apply Lemma 3.1 with κ = 1 + 1/ log x.The contribution of n ≤ x/2 to the error term is The contribution of n ≥ 2x to the error is Finally, if n ∈ (x/2, 2x), the contribution is where the second inequality follows e.g. by the argument in [Ten15, Cor.II.2.4].
As a special case of this corollary we have Corollary 3.3.Let χ be a Dirichlet character.We have x log x when s = 1 + 1/ log x, we see that perturbing the parameter T (appearing in the range of integration) by O(1) incurs an error of O(x log x/T ) which is absorbed in the existing error term.

3B Analytic continuation
Given a Dirichlet series G(s) associated with a multiplicative function g(n), which converges absolutely for s > 1 and does not vanish there, we define the kth root of G as for each positive integer k, where the logarithm is chosen so that arg G(s) → 0 as s → ∞.The function G 1/k is also a Dirichlet series, since Lemma 3.5.Let χ be a Dirichlet character modulo q.We have, for s > 1, for G which is analytic and nonvanishing in s > 1/4 and bounded in s Each g p is analytic in s > 0. Fix c > 0. For s ≥ c we have by Taylor expanding log(1 + x).In particular, if c = 1/4 + ε we have and so G may be extended to s > 1/4 via (3.2).As each g p is analytic in s > 1/4, and p≡3 mod 4 g p (s, χ) converges uniformly in s ≥ 1/4 + ε for every choice of ε > 0, it follows that log G is analytic in s > 1/4 and so is G.The formula for G(1/2, χ) for real χ follows from evaluating g p at s = 1/2 and observing χ 2 = χ 0 .
Assuming GRH for L(s, χ) where χ is nonprincipal, L(s, χ) 1/k may be analytically continued to the region and is nonzero there (because of trivial zeros of L(s, χ) we have to remove {σ : σ ≤ −1}).For χ principal we have a singularity at s = 1 and so L(s, χ) 1/k may be analytically continued to Hence, given χ which is nonprincipal and such that χχ −4 is nonprincipal, we have the following.Under GRH for χ, χχ −4 , χ 2 and χ 2 χ −4 , we may continue F (s, χ) analytically to if both χ 2 and χ 2 χ −4 are nonprincipal; otherwise we may continue it to

3C L-function estimates
We quote three classical bounds on L-functions from the book of Montgomery The following is a consequence of the functional equation.
These four lemmas are originally stated for primitive characters.However, if χ is induced from a primitive character ψ, then in s > 0 the ratio L(s, χ)/L(s, ψ) is equal to the finite Euler product p: χ(p)=0 (1 − ψ(p)/p s ).This product is bounded away from 0 and from ∞ when s ≥ ε, so we can convert results for L(s, ψ) to results for L(s, χ) as long as we restrict our attention to σ ≥ ε.

3D Contour choice
Let χ be a nonprincipal Dirichlet character modulo q.Fix c ∈ (0, 1/8) (say, c = 1/10).Let T 1.We want to use Cauchy's Integral Theorem to shift the vertical contour appearing in (3.1) to the left of s = 1/2 (namely to s = 1/2 − c), at the 'cost' of certain horizontal contributions.As we want to avoid zeros of L(s, χ)L(s, χχ −4 ) and poles and zeros of L(2s, χ 2 )/L(2s, χ 2 χ −4 ) (which by GRH can only occur at s = 1/2), we will use (truncated) Hankel loop contours to go around the relevant zeros and poles; the integrals over these loops will be the main contribution to our sum.It will also be convenient for 1/2 − iT and 1/2 + iT to avoid zeros of L(s, χ)L(s, χχ −4 ); this is easy due to Remark 3.4, showing that changing T by O(1) does not increase the error term arising from applying Perron's (truncated) formula.We replace the range be the imaginary parts of the zeros of L(s, χ)L(s, χχ −4 ) on σ = 1/2 with t ∈ (−T, T ) (without multiplicities), and, if either χ 2 or χ 2 χ −4 is principal, we include the number 0 (if it is not there already).Let r ∈ (0, 1) be a parameter that will tend to 0 later.Consider the contour where I 1 traverses the horizontal segment from right to left, I 2 traverses the horizontal segment from left to right, J j traverses the following vertical segment from its bottom point to the top: and finally each H ρ traverses the following truncated Hankel loop contour in an anticlockwise fashion: where in our case c = 1/10 and r = o(1/ log x).We refer the reader to Tenenbaum [Ten15, pp.179-180] for background on the Hankel contour and its truncated version.If r is small enough, the contour in (3.6) does not intersect itself.If both χ 2 and χ 2 χ −4 are nonprincipal characters and the corresponding L-functions satisfy GRH observe 4 L(2s, χ 2 )/L(2s, χ 2 χ −4 ) is analytic in s > 1/2 − 2c > 1/4 and so is F (s, χ) by Lemma 3.5.
If χ 2 is principal then χ 2 χ −4 cannot be principal.Similarly, if χ 2 χ −4 is principal then χ is a nonreal Dirichlet character of order 4 and χ 2 cannot be principal.In both cases, 4 L(2s, χ 2 )/L(2s, χ 2 χ −4 ) has an algebraic singularity at s = 1/2, which we avoid already as we inserted 0 to the list (3.5) if it is not there already.
In any case, by Cauchy's Integral Theorem, 1 2πi Lemma 3.10.Let χ be a nonprincipal Dirichlet character.Assume GRH holds for the following four characters: χ, χχ −4 , χ 2 , χ 2 χ −4 . (3.9) Let c ∈ (0, 1/8) be a fixed constant.Let T 1.We have The implied constant and A depend only on χ and c.We first treat I j , and concentrate on I 2 as the argument for I 1 is analogous.We have, using Lemma 3.5, It is now convenient to consider σ ≥ 1/2 and σ ≤ 1/2 separately.
If σ ≥ 1/2 we bound all the relevant L-functions using Lemmas 3.6 and 3.7, obtaining that this part of the integral contributes exp where A is a constant large enough depending on χ.For σ below 1/2 we first apply Lemma 3.9 to the Lfunctions of χ and χχ −4 to reduce to the situation where the real parts of the variables inside the L-functions are ≥ 1/2.Then we apply Lemmas 3.6 and 3.7 as before, obtaining that this part of the integral contributes It follows that We turn to the contribution of J j .We have where

Hankel calculus
In this section, H ρ is the Hankel contour described in (3.7), going around ρ in an anticlockwise fashion.Here the o(1) exponent goes to 0 as γ goes to ∞ (and it might depend on χ), and the implied constant is absolute.
Proof.Since we can write L(s, χ) as L(ρ, χ) plus an integral of L (z, χ) over a line segment connecting s and ρ, it follows that the maximum we try to bound is max |s−ρ|≤r, or s=σ+iγ with By Cauchy's integral formula, Lemma 3.6 and Lemma 3.9, implying the desired bound.
Proof.We first integrate by the x-variable and then take absolute values, obtaining that the integral is The result now follows from Lemma 4.2.
Proof.This is a variation on Lemma 4.2.If χ 2 χ −4 is principal we have , where the implied constant depends on χ only.Similarly, if χ 2 is principal and by Lemma 4.1.In both cases the integral is Let C q be the following positive constant, depending only on q: Lemma 4.5.Let χ be a nonprincipal Dirichlet character modulo q.Assume GRH for the characters in (3.9).If χ 2 is principal and where the implied constants depend only on χ.

Making the change of variables
where C q is defined in (4.1).The implied constants depend only on q.
The following is an ω-analogue of Lemmas 4.4 and 4.5.
Proof.The first part is a minor modification of the proof of Lemma 6.3.The second part is [Men20, Eq. (2.28)].
Lemma 6.4 holds for Ω in place of ω, with the only difference being a sign change in (6.4).

6B Proof of Theorem 2.1
We shall prove the theorem in the case of ω; the proof for Ω is analogous.Suppose a, b satisfy (a, q) = (b, q) = 1.Suppose the constant D q,a,b appearing in (2.9) is positive.Consider X 1 which will tend to ∞.By orthogonality of characters we write n≤x n≡a mod q ω(n) − n≤x n≡b mod q ω(n) = 1 φ(q) + o(1) .
As in the proof of Theorem 1.1, Chebyshev's inequality allows us to conclude the following.The probability that for a number x chosen uniformly at random from [X, 2X], n≤x, n≡a mod q ω(n) < n≤x, n≡b mod q ω(n) tends to 1 with X.This finishes the proof.

Table 1 :
for a table of the values of C 15,a,b and Table 2 for #{n ≤ 10 7 : S(n; 15, a) > S(n; 15, b)}/10 7 (in percentages).We omit pairs with a ≥ b due to symmetry and pairs a, b with a/b being a quadratic residue.Values of C 15,a,b .