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1 n 1 lim inf # p 6 x : > 0 > . (1.2) x + π(x) p 2 → ∞ 6 nXαp We were able to prove Conjecture 1 for rational α with small denominators and 1 also for all real α inside a small neighbourhood of the point 3 . So, our two main results are as follows: 6 1 Theorem 2. Conjecture 1 holds for all rational α 2 with denominators in the set 1, 2, 3, 4, 5, 6, 8, 12 . { } 1 6 6 6 Theorem 3. Conjecture 1 holds for all real α satisfying 3 2 10− α 1 6 − · +2 10− . 3 · To prove these two theorems, we are going to reduce the initial problem to the study of certain fixed random variable, using distribution of primes in arithmetic progressions and various methods of Fourier analysis and probability theory.
§ 2. Fourier expansion of L(α, p)
In this section we are going to prove the following simple result: Theorem 4. For any α and all large enough prime numbers p the equality
2πiαm 1 e− m L(α, p)= τ · − (2.1) p 2πim p m Z,m=0 ∈X 6 p 2πin/p n holds, where τ p· = e p is quadratic Gauss sum. n=0 Proof. Theorem is trivialP for α Z, so it is enough to assume that α is not ∈ an integer. Also, L(α, p) is a periodic function of α with period 1 and so is the right-hand side of our formula. Thus, we can also assume that 0 <α< 1. As periodic characteristic function χ ( x ) of the interval [0, α] is smooth everywhere [0,α] { } except for the discontinuity points, by Dini’s critertion for x 0, α (mod 1) we have 6≡ χ ( x )= χ (m)e2πimx, [0,α] { } F [0,α] m Z X∈ where
α 2πimx α if m =0 −2 χ[0,α](m)= e− dx = 1 e πiαm F 0 ( − if m =0. Z 2πim 6 LONG NONNEGATIVE SUMS OF LEGENDRE SYMBOLS 3
0 Therefore, due to the fact that p =0 and αp is not an integer for large enough p, the equality
p p n n L(α, p)= χ (n/p) = χ (m)e2πnm/p [0,α] p F [0,α] p n=0 n=0 m Z X X X∈ holds for p large enough. Changing the order of summation and using the fact that the sum of Legendre symbols over a complete system of residues is equal to 0, we get
2πiαm p 1 e− n L(α, p)= − e2πinm/p . 2πim p m=0 n=0 X6 X From multiplicativity of Legendre symbol, we easily obtain
p n m e2πinm/p = τ · , p p p n=0 X from which we get the desired result.
§ 3. Probabilistic reduction
Here we show that, roughly speaking, in our formula for L(α, p) one can replace all the Legendre symbols by the random multiplicative function. It turns out that it is possible to reduce certain properties of linear combinations of Legendre symbols to properties of random completely multiplicative functions f satisfying f(n)= 1 ± for all n N. ∈ Let us give a few definitions. The main object of our study is the random prime number: Definition 1. For ε = 1 and x > 5 by pε we denote the random variable ± x which is uniformly distributed among primes 6 x that are congruent to ε modulo 4. Next, we need to define the random multiplicative function:
Definition 2. Let X2,X3,X5,X7,X11 ... be a sequence of independent identi- cally distributed random variables which are distributed according to the Rademacher distribution and indexed by prime numbers. In other words, for any prime p we have 1 P(X =1)= P(X = 1) = p p − 2
For any natural number n we define Xn by the formula
νp(n) Xn = Xp , p Y where ν (n) is the largest k with pk n. Note that the product contains finite p | number of terms and also that Xab = XaXb for all a and b. 4 A.B. KALMYNIN
Using constructed random variables, we define certain random series. Definition 3. Let a be a sequence of complex numbers, x > 5 and ε = 1. { n} ± We define
+ ∞ a n L(a,x,ε)= n n pε n=1 x X and
+ ∞ a X L(a)= n n . n n=1 X In this section we will show that L(a) is often a rather good model for L(a,x,ε). But first of all, we need to show that L(a) is well-defined. Lemma 1. If the sequence a is bounded, then the series defining L(a) con- { n} verges almost surely. Proof. Assume that a 6 C for all n. Consider the fourth moment of the | n| weighted sum of Xn:
E( a X + ... + a X 4)= a a a a EX . | 1 1 n n| p q r s pqrs 6 p,q,r,sX n Using the boundedness of an and noticing that EXn =0 unless n is a square, in which case EXn =1, we get
4 4 E( a1X1 + ... + anXn ) 6 C 1. | | 2 pqrs=m p,q,r,sX6n Now, if pqrs = m2 and p,q,r,s 6 n then m 6 n2, so
4 4 2 E( a1X1 + ... + anXn ) 6 C τ4(m ). | | 2 mX6n As τ (m2) mε for any ε> 0 we obtain 4 ≪ E( a X + ... + a X 4) n2+ε. | 1 1 n n| ≪ Choosing ε =1/6 we get by Markov’s inequality
2+1/6 5/6 n 7/6 P( a X + ... + a X > n ) = n− . | 1 1 n n| ≪ n20/6 5/6 Hence, by Borel-Cantelli lemma we have a1X1 + ... + anXn = O(n ) almost surely. Using partial summation, we obtain the convergence of
a X n n . n > nX1 Now we are going to use some results on primes in arthmetic progressions to prove the following theorem LONG NONNEGATIVE SUMS OF LEGENDRE SYMBOLS 5
Theorem 5. Let a be a bounded sequence of real numbers such that the in- { n} equality
n max an √p ln p N p ≪ 6 nXN holds for all but at most o(π(x)) primes p 6 x as x + (i.e. abovemen- → ∞ tioned inequality is true for almost all primes). Then for ε = 1 random variables ± L(a,x,ε) converge to L(a) in distribution as x + . → ∞ Proof. The proof will be divided into several parts in which we will treat dif- ferent chunks of our series differently. We will formulate and use several lemmas concerning the distribution of prime numbers and the method of moments inside the proof. First of all, we split the sum in the definition of L(a,x,ε) into three sums as follows:
an n an n an n L(a,x,ε)= ε + ε + ε = 3 n px 3 2 n px 2 n px n6Xln x ln x
A(a,x,ε)+ B(a,x,ε)+ C(a,x,ε). We will prove that as x + the variables B(a,x,ε) and C(a,x,ε) both con- → ∞ verge to 0 in probability and then show that A(a,x,ε) converges to L(a) in distri- bution via the method of moments. Let us prove that the infinite sum C(a,x,ε) is usually small. This easily fol- lows from the conditions of Theorem 5 because with probability 1 o(1) we have n ε − an p √p ln p, so for almost all realizations p of px we get n6N ≪ P n n 2 a n d n6t an p n6√x ln x an p n = = + 2 2 n p P t −P √x ln x n>√Xx ln x √xZln2 x
n n6t an p √p ln p 1 dt = O 2 = O , P t2 √x ln x ln x √xZln2 x as p 6 x. Therefore, C(a,x,ε) converges to 0 in probability as x + . To → ∞ handle B(a,x,ε), let us estimate the expectation of B(a,x,ε)2. Notice the following n property of p : Lemma 2. For any nonzero integer a there is a Dirichlet character χa of mod- ulus at most 4 a such that for any odd integer n we have a = χ (n) and χ is | | n a a nonprincipal if and only if a is not a square. Proof. For any a there are b and c such that a = bc2, b is a fundamental discriminant and 2c is an integer. Now, for any odd integer n we have 6 A.B. KALMYNIN
a b c2 b = = χ (n). n n n n 0,2c b As b is a fundamental discriminant, n is a Dirichlet character to the modulus b (see [2], p. 53). Therefore a = b χ (n) = χ (n), where χ has modulus | | n n 0,2c a a at most 2 b c 6 4 b c2 =4a, which completes the proof. | | | | Now, the expectation of B(a,x,ε)2 equals
2 1 an n π(x, 4,ε) n p p6x ln3 x Due to nonnegativity of summands, we can sum over all odd numbers instead of primes and get 2 1 a n EB(a,x,ε)2 6 n = π(x, 4,ε) n d d6x ln3 x 1 a a nk 1 b m = n k = m , π(x, 4,ε) nk d π(x, 4,ε) m d d6x ln3 x where bm = anak. As an = O(1) for all n, we have bm = ln3 x 1 b EB(a,x,ε)2 6 | m| χ (d) π(x, 4,ε) m m ln6 x Now, if m is a square, then inner sum is trivially estimated by x, while if it isn’t a square, the P´olya-Vinogradov inequality gives us the bound O(√m ln m). Also, π(x, 4,ε) x , therefore ∼ 2ln x ln x τ(l2) τ(m) ln m E 2 B(a,x,ε) x 2 + . ≪ x 3 2 l 6 4 √m ln x τ(m) ln m x2/3. 6 4 √m ≪ ln x 1 EB(a,x,ε)2 , ≪ ln x which implies that B(a,x,ε) converges to 0 in probability as x + . → ∞ So, we are left with the shortest part of our sum. To prove that A(a,x,ε) converges to L(a) in distribution, we are going to use the moment method in the following form: Lemma 3 (Carleman’s criterion). Let ξn be a sequence of real random vari- ables such that for some random variable ξ and all natural numbers k the identity lim Eξk = Eξk n + n → ∞ holds. Assume that ξ satisfies Carleman’s criterion 2n 1/2n (Eξ )− =+ . ∞ > nX1 Then ξ converges to ξ in distribution for n + . n → ∞ Proof. See [1], p. 85. In the moment computation we will use the following classical result on primes in arithmetic progressions Lemma 4 (Siegel-Walfisz theorem). Let A> 0 be a fixed real number. Then there is a positive constant cA such that for any real x> 0 and nonprincipal Dirichlet character χ modulo q 6 (ln x)A the estimate cA√ln x π(x; χ)= χ(p)= O xe− 6 pXx holds. Proof. See [3], p. 138. As an immediate consequence, we obtain the following result on expectations of Legendre symbols: Corollary 1. For any positive integer k there is a positive constant ck such that for all 0 Proof. Indeed, this expectation can be rewritten as n 1 n 1 n 4n E = = + ε − . pε π(x, 4,ε) p 2π(x, 4,ε) p p x p6x p6x p ε X(mod 4) X ≡ Applying Lemma 2 and Lemma 4, we get the desired result. Using Corollary 1, one can easlily deduce the formula for k th moment of − A(a,x,ε). Note first that k an1 an2 ...ank n1 ...nk A(a,x,ε) = ε . 3 n1 ...nk px ni6Xln x Rearranging the summands according to the product of variables, we get τ (n; a, x) n A(a,x,ε)k = k , pε 3k n x n6Xln x where τk(n; a, x) = an1 ...ank . Also, one can easily show that 3 n1...nk=n,ni6ln x τ (n; a, x) τ (n) and τ (n;Pa, x) τ (n; a)= τ (n; a, ). Using the linearity of | k | ≪ k k → k k ∞ expectation and Corollary 1, we deduce that τ (n2; a, x) E k k √ A(a,x,ε) = 2 + O(exp( 0.5ck ln x)). k n − n6Xln x Thus, by dominated convergence theorem we obtain τ (n2; a) lim EA(a,x,ε)k = k . x + n2 → ∞ n X Now, using dyadic subdivision, one can prove that the k th moment of L(a) − exists and equals a ...a EX τ (n2; a) E k n1 nk n1...nk k L(a) = = 2 , n1 ...nk n n1,...,n n X k X because the expectation in this sum is only nonzero when n1 ...nk is a square. We now need to prove that L(a) satisfies Carleman’s condition. To do so, notice that τ (n2; a) 6 Ckτ (n2). Therefore, | k | k τ (n2) EL(a)2k 6 C2k 2k . n2 n X 2 As τ2k(n ) is multiplicative, we get 1 1 1 EL(a)2k 6 C2k + . 2 (1 p 1)2k (1 + p 1)2k p − − Y − Let us split this product in two parts. For primes p 6 k2 we use the inequality LONG NONNEGATIVE SUMS OF LEGENDRE SYMBOLS 9 1 1 1 1 + 6 , 2 (1 p 1)2k (1 + p 1)2k (1 p 1)2k − − − − − from which we deduce by Mertens’ third theorem that 1 1 1 1 6 6 2k 2k 1 2k + 1 2k 1 2k B (ln k) 2 2 (1 p− ) (1 + p− ) 2 (1 p− ) pY6k − pY6k − for some absolute constant B. Further, for p > k2 the relation 2 2k 4 2k 2k 2k 1 1 1 1+ p− + p− + ... + p− + = 2 4 2k 2 (1 p 1)2k (1 + p 1)2k (1 p 2)2k − − − − − holds. 2k 6 m Using the inequality m (2k) for all m, we obtain for k> 2 the estimate 1 1 1 1+8p 2k2 + 6 − 2 (1 p 1)2k (1 + p 1)2k (1 p 2)2k − − − − − Thus for all k> 2 we get 1 1 1 1+8p 2k2 6 − 1 2k + 1 2k 2 2k 2 2 (1 p− ) (1 + p− ) 2 (1 p− ) p>kY − p>kY − π2 2k 8k2 π2 2k 6 exp 2 . 6 2 p ≪ 6 p>kX Combining these bounds we obtain for some fixed B and C π2 2k EL(a)2k BC ln k . ≪ 6 This upper bound implies that 2k 1/2k 1 (EL(a) )− =+ . ≫ ln k ∞ > > kX1 kX3 Therefore L(a) satisfies Carleman’s condition and this completes the proof. Theorem 5 can be used in very different settings, but we are going to use the following corollary: 2πinαm Corollary 2. If an = m λme for some finite sequence of complex λm and real α and a is real for all n, then L(a,x,ε) converges to L(a) in distribution. m n P Proof. Obviously, an is bounded. Therefore, it is enough to check that for any real α the estimate n e2πiαn √p ln p p ≪ 6 nXN 10 A.B. KALMYNIN holds for almost all primes p. To show that this is indeed the case, note that if α is rational and nonzero then αp 1 for p large enough and if α is irrational || || ≫ then αpn is uniformly distributed modulo 1 (see [2; p.489, Theorem 21.3]), so for almost all p the inequality αp > 1 holds. Using the Fourier expansion for the || || ln p Legendre symbol, we get for almost all p p 1 p 1 − − 2πiαn n 1 1 1 e = exp(2πin(α a/p)) α a/p − p − ≪ p || − || ≪ 6 τ · a=1 6 √ a=1 nXN p X nXN X p/2 1 p √p ln p, ≪ √p a +1/ ln p ≪ a= p/2 X− | | which completes the proof for α = 0. If α = 0, then the desired bound is just 6 the P´olya-Vinogradov inequality. Corollary 2 has a very useful implication on our main problem. Namely, from Section 2 we deduce that we need to study the distribution of L(a±, x, 1) with ± an± being an exponential polynomial of variable nα. We are also going to use the following well-known formula for the Gauss sum: Lemma 5. Let p be an odd prime number. Then we have √p if p 1 (mod 4) τ · = ≡ p (i√p if p 3 (mod 4). ≡ Proof. See [2], p. 49. From this we get the following: + Corollary 3. Let α be a real number, a (α) = sin 2πnα and a−(α)=1 n n − cos2πnα. Let c(α) be a relative lower density of primes p, satisfying L(α, p) > 0, + + c (α) and c−(α) be the probabilities of positivity of random variables L(a (α)) and L(a−(α)). Then we have c+(α)+ c (α) c(α) > − . 2 + In particular, if c (α)+ c−(α) > 1, then Conjecture 1 is true for α. Proof. First, it is enough to only consider large enough odd prime numbers, because we can ignore any finite number of primes. Now, if p 1 (mod 4) then we ≡ have by Theorem 4 2πiαm 1 e− m L(α, p)= τ · − = p 2πim p m Z,m=0 ∈X 6 2πiαm 2πiαm 1 e− 1 e m √p sin 2πmα m = √p − − = , 2πim − 2πim p π m p m>0 m>0 X X LONG NONNEGATIVE SUMS OF LEGENDRE SYMBOLS 11 m 1 m m 6 because −p = −p p = p . Therefore, the proportion of primes p x with p 1 (modp) with L(α, p) >0 among all primes p 6 x is ≡ π(x;4, 1) P(L(a+, x, 1) > 0) P(L(a+, x, 1) > 0) π(x) ∼ 2 as x + . Now, as L(a+, x, 1) converges to L(a+) in distribution and the → ∞ subset R R is open, we have >0 ⊂ lim inf P(L(a+, x, 1) > 0) > lim inf P(L(a+, x, 1) > 0) > P(L(a+) > 0) = c+(α). x + x + → ∞ → ∞ If p 3 (mod 4), then we have similarly ≡ 2πiαm 1 e− m L(α, p)= τ · − = p 2πim p m Z,m=0 ∈X 6 2πiαm 2πiαm 1 e− 1 e m √p 1 cos2πmα m = i√p − + − = − . 2πim 2πim p π m p m>0 m>0 X X Hence, the proportion of primes p 6 x with p 3 (mod p) with L(α, p) > 0 ≡ among all primes p 6 x is π(x;4, 3) P(L(a−, x, 1) > 0) P(L(a−, x, 1) > 0) − . π(x) − ∼ 2 And we also have lim inf P(L(a−, x, 1) > 0) > lim inf P(L(a−, x, 1) > 0) > P(L(a−) > 0) = c−(α). x + x + → ∞ → ∞ Therefore, we obtain + # p 6 x : L(α, p) > 0 c (α)+ c−(α) lim inf { } > , x + → ∞ π(x) 2 as needed. § 4. Rational α with small denominators In this section we are going to prove Theorem 2. In other words, we are going to prove Conjecture 1 for the following values of α: 1 1 1 1 1 3 1 5 1 2 α =0, , , , , , , , , , . 2 3 4 6 8 8 12 12 5 5 All the denominators of these numbers, except for 5, are precisely the numbers n such that the group (Z/nZ)∗ has exponent 1 or 2, i.e. such that a2 1 (mod n) for ≡ all a coprime to n. The reason for such a choice of denominators is that for these n all the Dirichlet characters modulo n are real-valued and so we don’t need to 1 2 consider any complex Euler products. On the other hand, in the case of α = 5 or 5 12 A.B. KALMYNIN we need to compute arguments of certain complex-valued Euler products, which is going to make things more complicated. Probably, the list of n such that Conjecture 1 follows from conditions nα Z and α < 1 can be expanded in a way similar to ∈ 2 what is presented in this section, but we don’t know if it works for all rational α. For a natural number m the function χ ( ) will be the principal character 0,m · modulo m. If χ is a Dirichlet character and β is not an integer then we are going n to define χ(β) to be equal 0. So, for example, χ0,2 2 is a 4-periodic function with values 0, 1, 0, 0, 0, 1,... Proof of Theorem 2. Let us start with a few simple cases. If α =0 then our conjecture is trivial, as we always have L(0,p)=0, so c(0) = 1. 1 If α = 2 , then + an (α) = sin πn =0 for all n and 2 if n is odd an−(α)=1 cos πn = − (0 otherwise From these formulas we get L(a+)=0 for all possible realizations of (X ). Therefore, for all primes p 1 (mod 4) n ≡ we have L(1/2,p)=0. Indeed, in the previous section we learned that for p 1 π ≡+ (mod 4) the quantity √p L(α, p) is a particular realization of random variable L(a ), while for p 3 (mod 4) it is a realization of random variable L(a−). From our ≡ formulas we also get 1 χ0,2(n)Xn χ0,2(p)Xp − L(a−)=2 =2 1 > 0, n − p n>0 p X Y due to multiplicativity of χ0,2(n)Xn. Therefore, for all primes p we have L(1/2,p) > 0, so c(1/2)=1. 1 Now, for α = 3 one can easily check that 2πn √3 n a+(1/3) = sin = (4.1) n 3 2 3 and 2πn 3 a−(1/3)=1 cos = χ (n). (4.2) n − 3 2 0,3 and we again get nonnegativity of all realizations from the Euler products for corresponding series. Hence in this case we again have c(α)= c(1/3)=1. 1 The same is true for α = 4 , because in that case we get πn 4 a+(1/4) = sin = − = χ (n) n 2 n 4 LONG NONNEGATIVE SUMS OF LEGENDRE SYMBOLS 13 and πn n a−(1/4)=1 cos = χ (n)+2χ , n − 2 0,2 0,2 2 therefore c(1/4)=1, because 1 χ (p)X − L(a+)= 1 4 p > 0 − p p Y and 1 1 χ2,0(p)Xp − χ0,2(p)Xp − L(a−)= 1 + X 1 = − p 2 − p p p Y Y 1 X − (1 + X ) 1 p > 0. 2 − p p>2 Y The factor X2 appears here because for any χ and any d we have χ(n/d)X χ(n)X X χ(n)X n = nd = d n . n nd d n n n n X X X 1 For α = 6 we once again obtain c(1/6)=1, but for a more complicated reasons. Here we have πn a+(α) = sin . n 3 The values of a+ are √3 , √3 , 0, √3 , √3 , 0,... (the sequence is 6-periodic) so n 2 − 2 − 2 2 we get √3 n/2 a+(α)= χ (n)+ . n 2 6 3 Here χ6(n) = 1 if n 1 (mod 6) and 0 otherwise. Note also that χ6(n) = n ± ≡ ± χ0,2(n) 3 . Therefore, n/2 n X √3 χ0,2(n) Xn 3 n L(a+)= 3 + = 2 n n n n X X 1 p − √3 Xp (1 + X ) 1 3 > 0, 2 2 − p p ! Y so we again have L(α, p) > 0 for all p 1 (mod 4). 1 3 3 1 ≡ Values of an− are 2 , 2 , 2, 2 , 2 , 0 ... and this function can be expanded as follows: πn n 1 n a− =1 cos =2χ + χ (n)+ χ . n − 3 0,2 3 2 0,3 0,3 2 Nonnegativity is not obvious from this formula, so let us wor k with the Euler products 14 A.B. KALMYNIN an− 2X3 χ0,2(n)Xn 1+ X2 χ0,3(n)Xn L(a−)= = + = n 3 n 2 n n n n X X X 2X X 1+ X X X 1+ X + X X X X = 3 1 2 + 2 1 3 n = 2 3 − 2 3 n > 0. 3 − 2 2 − 3 n 2 n n n X X This calculation shows that c(1/6)=1. Now we are going to consider a different set of values α for which we don’t have c(α)=1. 1 + πn If α = 8 then the first 8 values of an (1/8) = sin 4 are equal to √2 √2 √2 √2 , 1, , 0, , 1, , 0, 2 2 − 2 − − 2 √2 2 n which is equal to 2 −n + χ4 2 thus we obtain 2 √2 − Xn X χ (n)X L(a+)= n + 2 4 n 2 n 2 n n n X X + 1 + > and hence c (1/8) > 2 , because for X2 = +1 we have L(a ) 0 and for X = 1 our random variable is positive with positive probability, because 2 − √2 1 1 (√2 1)π2 E(L(a+) X = 1) = − = − > 0. | 2 − 2 n2 18 X2∤n On the other hand, the first 8 values of an−(1/8) are √2 √2 √2 √2 1 , 1, 1+ , 2, 1+ , 1, 1 , 0 − 2 2 2 − 2 and one can check that we have √2 2 a−(1/8) = χ (n)+ χ (n/2)+2χ (n/4) . n 0,2 0,2 0,2 − 2 n Therefore 2 3 X2 χ0,2(n)Xn √2 n Xn L(a−)= + > 2 2 2 − 2 n n n X X 2 χ (n)X √2 Xn √2 > 0,2 n n = F G. n − 2 n − 2 n n X X Now, let us notice, that the distribution of (λ(n)Xn), where λ(n) is the Liouville’s function, coincides with the distribution of (X ), as X are also independent and n − p have Rademacher’s distribution. On the other hand, if we replace Xn by λ(n)Xn, F and G will transform into 1 1 λ(n)χ (n)X X − 1 − X π2 0,2 n = 1+ p = 1 1 p = n p − p2 − p 9F n p>2 p>2 X Y Y LONG NONNEGATIVE SUMS OF LEGENDRE SYMBOLS 15 2 and (by the same argument) π , respectively. Also, ln F ln G has continuous 9G − distribution (because characteristic function goes to 0 at infinity), so P(L(a−) > > 1 √2 > 0) 2 , because F 2 G 0 holds either for Xn or for λ(n)Xn, so we get 1 − c(1/8) > 2 . 3 As for the second case of denominator 8, i.e. α = 8 , using a3±n(1/8) = an±(3/8) we get 2 √2 − Xn X χ (n)X L(a+(3/8)) = n 2 4 n , 2 n − 2 n n n X X + + 1 so c (3/8) = c (1/8) > 2 . Also, in the formula for L(a−) we will get +G instead of G, so that − √2 L(a−) > F + G > 0. 2 > 3 Hence we get c−(3/8)=1 and c(3/8) 4 . 1 The last two cases without any complex Dirichlet characters are α = 12 and 5 α = 12 . 1 + πn Let us start with α = 12 . For an = sin 6 we get the first twelve values 1 √3 √3 1 1 √3 √3 1 , , 1, , , 0, , , 1, , , 0, 2 2 2 2 −2 − 2 − − 2 −2 find the formula 1 √3 n/2 √3 n/4 a+(1/12) = χ (n)χ (n)+ χ (n/2)+ χ (n/3)+ n 2 4 0,3 2 3 0,2 4 2 3 and get 1+ X χ (n)X √3(X + 1) χ (n)X L(a+)= 3 4 n + 2 3 n , 2 n 4 n n n X X thus, L(1/12,p) > 0 for all p 1 (mod 4). ≡ For cosines we compute πn √3 1 3 √3 √3 3 1 √3 1 cos =1 , , 1, , 1+ , 2, 1+ , , 1, , 1 , 0, − 6 − 2 2 2 2 2 2 2 − 2 so that √3 12 1 3 a− = + χ (n)+ χ (n/2)+ χ (n/4)+2χ (n/6). n − 2 n 0,2 2 0,6 2 0,3 0,2 Observe that an− is always nonnegative, so that we have EL(a−) > 0 and c−(1/12) > 1 0, which results in c(1/12) > 2 . 5 + When α = 12 we apply the formula an±(5/12) = a5n(1/12), and for a we get 16 A.B. KALMYNIN 1+ X χ (n)X √3(X + 1) χ (n)X L(a+)= 3 4 n 2 3 n , 2 n − 4 n n n X X so that L(a+) > 0 for almost all X with X = 1 and X = 1, therefore n 2 − 3 c+(5/12) > 1 . On the other hand, we have 12 = 1, so 4 5 − √3 12 1 3 a− = + χ (n)+ χ (n/2)+ χ (n/4)+2χ (n/6), n 2 n 0,2 2 0,6 2 0,3 0,2 which gives 12 1 Xn √3 n Xn L(a−)= (1 + (1 X )(1 X )) + , 4 − 2 − 3 n 2 n n n X X + > 1+1/4 5 so that c (5/12) = 1 and c(5/12) 2 = 8 . Now, for the last case, where we have to deal with some complex characters. If 1 2πn α = , then the function a− =1 cos takes values 5 n − 5 5 √5 5+ √5 5+ √5 5 √5 − , , , − , 0 4 4 4 4 and we get 5 √5 n 2 χ0,5(n) 2 5 Xn 5 √5 L(a−)= − = H T, n 2 − 2 n X where 1 χ (n)X X − H = 0,5 n = 1 p n − p n p=5 X Y6 and 1 n n X − Xn p p T = 5 = 1 . n − p n p=5 X Y6 2 2 4π 4π > 1 The map Xn λ(n)Xn transforms H into 25H and T into 25T , so c−(1/5) 2 . 7→ 2πn Expressions for sin 5 involve nested radicals, which gives us the formula 2πn A iB A + iB sin = − κ(n)+ κ(n), 5 2 2 where κ is a Dirichlet character modulo 5 with κ(2) = i and 5+ √5 5 √5 A = ,B = − . s 8 s 8 Therefore, the inequality L(a+) > 0 is equivalent to LONG NONNEGATIVE SUMS OF LEGENDRE SYMBOLS 17 κ(n)X Re (A iB) n > 0 − n n ! X Now, A + iB = √A2 + B2eiϕ with ϕ = arctan B 0.553 and our inequality A ≈ takes the form κ(n)X cos ϕ + arg n > 0. − n n !! X The sum inside cosine can be expanded into the Euler product: 1 κ(n)X κ(p)X − n = 1 p . n − p n p X Y Obviously, if κ(p) is real, then p gives no contribution to the sum, and if κ(p) is not real (which is equivalent to p 2 (mod 5)) then ≡± 1 1 κ(p)X − exp(κ(p)Xp arctan ) 1 p = p . − p 2 1+ p− From this we deduce the formula p κ(n)X n = Meiξ, n n X where M is an almost surely positive real random variable and κ(p) 1 ξ = X arctan , i p p p 2 (mod5) ≡± X which is also real. Therefore, our desired inequality takes form cos(ξ ϕ) > 0. − Notice that if it is not true, then we should have ξ ϕ > π , in which case | − | 2 ξ > π ϕ. | | 2 − By Chebyshev inequality, probability of this event is at most Eξ2 0.3536 1 6 2 < , π ϕ 1.077 3 2 − because 1 Eξ2 = arctan2 0.35355. p ≈ p 2 (mod5) ≡± X + > 2 > 7 1 From this we obtain c (1/5) 3 and so c(1/5) 12 > 2 , as needed. 2 If α = 5 , then the proof is easier, because in this case we have 4πn 5 √5 n a− =1 cos = χ (n)+ , n − 5 2 0,5 2 5 18 A.B. KALMYNIN thus 5 √5 L(a−)= H + T > 0, 2 2 2 > so for all p 3 (mod 4) we have L( 5 ,p) 0. On the other hand, the set of p 1 ≡ 2 ≡ (mod 4) with L( 5 ,p) > 0 has a positive lower density, because we have 2 sin 4πn 5 √5 n EL(a+)= 5 = − 5 > 0. n2 s 8 n2 n n>0 X X 1 From this we get c(2/5) > 2 , which concludes our proof. 1 § 5. Values of α close to 3 In the previous section we proved Conjecture 1 for several special values of α, using expansion with Dirichlet characters and certain symmetry considerations. Here we are going to use a different approach to the main conjecture. One can notice that all the random variables that we constructed are linear combinations of a fixed sequence of random variables Xn with coefficients that are smooth functions of a parameter α. Using this observation, we are going to prove Theorem 3. Let us define certain class of random variables that appears naturally when we study L(a±). Definition 4. Let σ2 > 0. We say that random variable lies in the class L(σ2), if there is an infinite sequence of real numbers a1,a2,... with 2 2 ai 6 σ . i X and a sequence κ1,κ2,... of independent Rademacher random variables such that η = aiκi. i X The next lemma will allow us to control the tail of distribution from the class L(σ2). Lemma 6. Let η be in L(σ2). Then for any T > 0 we have T 2 P(η > T ) 6 exp . −2σ2 Proof. Consider the moment generating function of η: F (t)= Eetη. It is easy to see that for any positive t the inequality tT P(η > T ) 6 e− F (t) is true. Let us show that F (t) 6 exp(t2σ2/2). Indeed, η lies in L(σ2), so for some ai and κi we have LONG NONNEGATIVE SUMS OF LEGENDRE SYMBOLS 19 ait ait t P a κ ta κ e− + e F (t)= Ee i i i = Ee i i = . 2 i i Y Y The expectation of exponent equals the product of expectations due to independence of κi. Note now that for any real t the inequality 2 2 t t t 2 4 2 e + e t t t 2 2 − =1+ + + ... 6 1+ + + ... = et /2 2 2 4! 2 2! holds. The upper bound is true due to the inequality (2n)! > 2nn! = (2n)!!. Therefore, for any real t we have 2 2 2 2 F (t) 6 eai t /2 6 eσ t /2, i Y from this we get 2 2 tT σ t /2 P(η > T ) 6 e− e . Choosing t = T/σ2, we prove the desired inequality. From this lemma we immediately deduce the bound for probability of negativity of random variables that are close to exp(η) for some η L(σ2). ∈ Corollary 4. Assume that real random variables X and Y satisfy X = eη for some η L(σ2) and ∈ E(X Y )2 6 D. − Then for any 0 ln2 u D P(Y 6 0) 6 exp + . − 8σ2 u Proof. Indeed, if Y 6 0 then either (X Y )2 > u or X2 6 u. Probability of the D − first event is at most u due to Markov’s inequality. On the other hand, the second event implies that η 6 ln u , therefore η > ln u . Notice now that if η L(σ2) − 2 − 2 ∈ then the same is true for η, as one can choose a instead of a in the defining − − i i formula. Thus, Lemma 6 implies that the probability of second event is at most 2 ln u exp 2 . This proves the desired estimate. − 8σ Let us show now that random variables L(a±(1/3)) are proportional to exponents of certain random variables from the class L(σ2). 2 2 Lemma 7. For σ =0.395 there are η1 and η2 from L(σ ) such that π L(a+(1/3)) = eη1 √3 and π η2 L(a−(1/3)) = e . 3 20 A.B. KALMYNIN Proof. Due to the formulas (4.1) and (4.2) we have 1 3 X χ (n) 3 X − L(a+(1/3)) = n 0,3 = 1 p 2 n 2 − p n>0 p=3 X Y6 and 1 n p − √3 3 Xn √3 Xp 3 ) L(a−(1/3)) = = 1 2 n 2 − p n>0 p=3 ! X Y6 Let us observe that for any ε = 1 the identities ± ε ε 1 1 1+ =1 − p p − p2 and 1 ε ε ε − p 1 1 1+ = − − p p p +1 hold. Multiplying this two equalities and taking a square root, we deduce ε p 1 ε/2 1 1/2 1 = − 1 . − p p +1 − p2 p If we now choose ε to be Xp and Xp 3 , we obtain the relations η1 L(a−(1/3)) = A1e and + η2 L(a (1/3)) = A2e , where 3 1 1/2 3 π2 3 4π2 π A1 = 1 = (1 1/9) = = , 2 − p2 2 6 − 2 27 √3 p=3 r r Y6 A1 π A2 = = , √3 3 1 p 1 η = ln − X 1 2 p +1 p p=3 X6 and 1 p 1 p η = ln − X . 2 2 p +1 3 p p=3 X6 Direct computation shows that the inequality 1 p 1 ln2 − < 0.395, 4 p +1 p=3 X6 is true, which concludes the proof of lemma. LONG NONNEGATIVE SUMS OF LEGENDRE SYMBOLS 21 1 Now we are going to show that if α is close to 3 then random variables L(a±(α)) and L(a±(1/3)) are in some sense close to each other. But first let us prove two auxiliary propostitions: Lemma 8. Let τ(n)= 1 be the divisor function. Then for any complex s with d n | Re s> 1 the equality P + ∞ τ(n2) ζ(s)3 = ns ζ(2s) n=1 X holds. Proof. Due to multiplicativity of τ(n2), we have + 2 2s ∞ τ(n ) s 2s 1 p− = 1+3p− +5p− + ... = − , ns (1 p s)3 n=1 p p − X Y Y − which gives the desired equality. Using this lemma, we will prove the following bound for L2 distance between two − random variables of certain type. Lemma 9. Let f : R R be a function with Lipschitz constant L such that → f(x) 6 C for any real x. Then for any pair of real α and β we have | | E(L(f(α)) L(f(β)))2 6 92 α β 2/3L2/3C4/3, − | − | where for any real γ the sequence fn(γ) is defined by formula fn(γ)= f(nγ) and L(f(γ)) then given by Definition 3 for an = fn(γ). Proof. By the definition of L(f(α)) L(f(β)), we have − (f(nα) f(nβ))(f(mα) f(mβ)) E(L(f(α)) L(f(β)))2 = − − E(X X ). − nm n m n,m>0 X 2 Let us notice that E(XnXm)=1 if nm = d for some integer d and 0 otherwise. Splitting the resulting sum according to range of d into d > A and d 6 A parts, we get (f(nα) f(nβ))(f(mα) f(mβ)) E(L(f(α)) L(f(β)))2 = − − + − 2 nm nm=Xd ,d6A (f(nα) f(nβ))(f(mα) f(mβ)) + − − . 2 nm nm=Xd ,d>A Every summand of the first sum is at most L2 α β 2 due to the Lipschitz 2 | − | 4C property, while in the second sum every summand is bounded by d2 . Furthermore, any d corresponds to exactly τ(d2) summands. From this we obtain the inequality 22 A.B. KALMYNIN τ(d2) E(L(f(α)) L(f(β)))2 6 α β 2L2 τ(d2)+4C2 . − | − | d2 6 dXA d>AX To get a more convenient estimate, we multiply every summand of the first sum by (A/d)4/3 and each summand in the second sum by (d/A)2/3. Obviously, both sums will not decrease. Thus, τ(n2) τ(d2) 6 A4/3 = A4/3s (A), n4/3 1 6 6 dXA dXA due to Lemma 8. Similarly, 2 2 τ(d ) 2/3 τ(d ) 2/3 6 A− = A− s (A). d2 d4/3 2 d>AX d>AX 3 ζ(4/3) Observing that s1(A)+ s2(A)= ζ(8/3) by lemma 8, we get for all positive A 2 2 2 4/3 2 2/3 E(L(f(α)) L(f(β))) 6 α β L A s (A)+4C A− s (A) 6 − | − | 1 2 3 ζ(4/3) 2 2 4/3 2 2/3 6 max( α β L A , 4C A− ) ζ(8/3) | − | 3 1 1 ζ(4/3) 4/3 Choosing A = 2CL− α β − and using relation 2 < 92, we obtain the | − | ζ(8/3) desired estimate. Applying Lemma 9 to the functions f(x) = sin(2πx) and f(x) = cos(2πx) with constants C =1 and L =2π, we obtain the inequality 2 2/3 E(L(a±(α)) L(a±(β))) 6 313.3 α β . (5.1) − | − | Now we have enough instruments to prove Theorem 2. Proof of Theorem 2. Due to Lemma 5 and inequality 5.1, the estimates √ η1 3 2 3 2 926.9 2/3 2/3 E(e L(a−(α))) = E(L(a−(1/3)) L(a−(α))) 6 α 1/3 6 94 α 1/3 − π π2 − π2 | − | | − | and 3 9 2780.7 E(eη2 L(a+(α)))2 = E(L(a+(1/3)) L(a+(α)))2 6 α 1/3 2/3 6 282 α 1/3 2/3 −π π2 − π2 | − | | − | hold. 6 From this for α 1/3 < 2 10− we deduce, using Corollary 4, that the probability | − | · of negativity of L(a−(α)) satisfies for any 0 ln2 u 0.015 P(L(a−(α)) 6 0) 6 exp + − 3.16 u and similarly LONG NONNEGATIVE SUMS OF LEGENDRE SYMBOLS 23 ln2 u 0.0447 P(L(a+(α) 6 0)) 6 exp + . − 3.16 u Computation shows that the minimal values of resulting expressions are attained at the points u 0.0756 and u+ 0.12957. Values of these expressions in − ≈ ≈ corresponding points are less than or equal to 0.32 and 0.612 respectively. From this we obtain the inequality 0.32+0.612 c(α) > 1 =0.534 − 2 6 in the desired range. Therefore, if α lies inside 2 10− -neighbourhood of 1/3, then · the sum L(α, p) is nonnegative for at least 53.4% of all primes, which concludes the proof. §6. Conclusion and open problems In this paper we were able to reduce Conjecture 1 to the study of one fixed random variable and partially prove the conjecture. Although the progress we made simplifies the original problem, there are still a lot of questions one may ask even beyond the Conjecture 1. For example, is it true, that the lim inf in 1.2 can be replaced by lim? If so, what are the properties of this limit as a function of α? Also, + from section 4 we see that sometimes probability of events L(a−)=0 or L(a )=0 + is nonzero: for example, if X2 = X3 = 1 then L(a (1/6)) = L(a−(1/6)) = 0. Is it + − possible to describe all α with P(L(a (α)) = 0)+P(L(a−(α)) = 0) > 0 and is this set finite? On the other hand, one might also ask about a computational perspectives of our results. As we use Siegel-Walfisz theorem in the proof of Theorem 5, the proof provides no method to estimate the rate of convergence of resulting distribution. It is probably possible to overcome this obstacle with the help of Page theorems but 1 we don’t know if this can be useful for numerical verification of inequality c(α) > 2 . References [1] Akhiezer, N. I. ¾The Classical Moment Problem and Some Related Questions in Anal- ysis¿, Oliver & Boyd, 1965 [2] Iwaniec H. , Kowalski E. , ¾Analytic Number Theory¿, American Mathematical So- ciety Colloquium Publications, vol. 53, American Mathematical Society, Providence, RI, 2004. [3] Karatsuba A. A. , ¾Basic Analytic Number Theory¿, Springer-Verlag Berlin Heidel- berg, 1993 [4] Schinzel A. Urbanowicz J. Van Wamelen P. ¾Class Numbers and Short Sums of Kronecker Symbols¿, J. Number Theory, 78 (1999), 62-84. A. B. Kalmynin Steklov Mathematical Institute of Russian Academy of Sciences, Gubkina, 8, Moscow, Russia E-mail: [email protected]