A. B. Kalmynin Long nonnegative sums of Legendre symbols Abstract. For 0 6 α < 1 and prime number p, let L(α, p) be the sum of the first [αp] values of Legendre symbol modulo p. We study positivity of 1 −6 1 L(α, p) and prove that for |α − 3 | < 2 · 10 and for rational α 6 2 with denominators in the set {1, 2, 3, 4, 5, 6, 8, 12} the inequality L(α, p) > 0 holds for majority of primes. § 1. Introduction n Let p be an odd prime number and χp(n)= p be the Legendre symbol modulo p. It is well-known that the sum n p nX6p/2 is always nonnegative. In other words, there are at least as many quadratic residues as nonresidues modulo p below p/2. More precisely, Dirichlet proved the following formula for this sum Theorem 1. Let p be an odd prime number. Then the equality n 2 2 h( p) if p 3 (mod 4) = − p − ≡ p (0 if p 1 (mod 4) nX6p/2 ≡ holds, where h( p) is the class number of the number field Q(√ p). − − Other cases of connection between character sums of this type and class numbers of quadratic fields are given in [4]. For example, the sum still will be nonnega- arXiv:1911.10634v2 [math.NT] 30 Jun 2021 tive if we replace p/2 by p/3 or p/4. This leads us to the general question about nonnegativity of the sum of length αp for any real number α. Let us define n L(α, p)= . p 6 nXαp 6 1 Numerical evidence suggests that for any α 2 most primes satisfy the inequality L(α, p) > 0. (1.1) 2 > For example, among first 1000 primes there are 896 with L( 5 ,p) 0, 917 with 3 > 1 > 1 > 1 > L( 8 ,p) 0, 884 with L( 12 ,p) 0, 812 with L( 2π ,p) 0 and 937 with L( e ,p) 0. For 10000 these numbers are 8915, 9122, 8799, 8019 and 9340, respectively, and we The work is supported by the Russian Science Foundation under grant №19-11-0001. ○c A.B. Kalmynin, 2021 2 A.B. KALMYNIN get 89041, 91036, 87868, 79784 and 93260 for the first 100000 prime numbers. As we can see, for all our choices of parameter α the proportion of p with nonnegative L(α, p) seems to be even more than 75%. Based on that, let us formulate our main conjecture: 6 6 1 Conjecture 1. For all 0 α 2 the lower asymptotic density of primes that 1 satisfy 1.1 is at least 2 . In other words, the inequality 1.1 holds for majority of prime numbers: 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<nX6√x ln x n>√Xx ln x A(a,x,ε)+ B(a,x,ε)+ C(a,x,ε).