Distribution of Residues Modulo $ P $ Using the Dirichlet's Class Number Formula

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1 Theorem 1. Let p be an odd prime. If p 3 (mod 4), then for any ǫ with 0 <ǫ< 2 , we have ≡ p 1 1 Q(p,S ) − p 2 −ǫ. 2 − 4 ≫ǫ When the prime p 1 (mod 4), we have ≡ p 1 Q(p,S )= − . 2 4 Corollary 1.1. Let p be an odd prime and let be the set of all odd integers in [1,p 1]. If R = N(p,S ) or R = Q(p, ), then for any Oǫ with 0 <ǫ< 1 , we have − 2 O 2 p 1 1 − R p 2 −ǫ, if p 3 (mod4). 4 − ≫ǫ ≡ When the prime p 1 (mod 4), we have ≡ p 1 R = − . 4 1 Theorem 2. Let p be an odd prime. If p 1, 11 (mod 12), then for any ǫ with 0 <ǫ< 2 , we have ≡ p 1 1 Q(p,S ) − p 2 −ǫ. 3 − 6 ≫ǫ When p 5, 7 (mod 12), in this method, we do not get any ﬁner information other than in (1).≡ Corollary 1.2. Let p be an odd prime. If p 1, 11 (mod 12), then for any ǫ with 1 ≡ 0 <ǫ< 2 , we have p 1 1 − N(p,S ) p 2 −ǫ. 6 − 3 ≫ǫ Theorem 3. Let p be an odd prime. Then, for p 3 (mod 8), we have ≡ 1 p 1 Q(p,S )= − . 4 2 4 1 Also, for any 0 <ǫ< 2 , we have p 1 1 Q(p,S ) − p 2 −ǫ, if p 1 (mod4), 4 − 8 ≫ǫ ≡ and 1 p 1 1 Q(p,S ) − p 2 −ǫ; if p 7 (mod8). 4 − 2 4 ≫ǫ ≡ Corollary 1.3. Let p be an odd prime. Then, for p 3 (mod 8), we have ≡ 1 p 1 N(p,S )= − . 4 2 4 1 Also, for any 0 <ǫ< 2 , we have p 1 1 − N(p,S ) p 2 −ǫ; if p 1 (mod4), 8 − 4 ≫ǫ ≡ and 1 p 1 1 − N(p,S ) p 2 −ǫ; if p 7 (mod8). 2 4 − 4 ≫ǫ ≡ DISTRIBUTION OF RESIDUES MODULO p USING THE DIRICHLET’S CLASS NUMBER FORMULA3

Using Theorems 1 and 3, we conclude the following corollary. Corollary 1.4. Let p be an odd prime such that p 3 (mod 8). Then for any ǫ with 1 ≡ 0 <ǫ< 2 , we have 1 p 1 1 Q(p,S S ) − p 2 −ǫ. 2\ 4 − 2 4 ≫ǫ

2. Preliminaries In this section, we shall state many useful results as follows. Theorem 4. (Polya-Vinogradov) Let p be any odd prime and χ be a non-principal Dirich- let character modulo p. Then, for any integers 0 M < N p 1, we have ≤ ≤ − N χ(m) √p log p. ≤ mX=M

Let us deﬁne the following counting functions as follows. Let 1 x (2) f(x)= 1+ for all x (Z/pZ)∗ 2 p ∈ and 1 x (3) g(x)= 1 for all x (Z/pZ)∗ 2 − p ∈ where · is the Legendre symbol. Then, we have p 1; if x is a quadratic residue (mod p), f(x)= 0; otherwise. and 1; if x is a quadratic nonresidue (mod p), g(x)= 0; otherwise. In the following lemma, we prove the “expected” result.

Lemma 1. For an integer k 1 and an odd prime p, let Sk = kI where I is the interval I = 1, 2,..., [(p 1)/k] . Then≥ { − } (p−1)/k 1 p 1 1 k m (4) Q(p,S )= − + k 2 k 2 p p mX=1 and hence 1 p 1 Q(p,S )= − + O(√p log p). k 2 k

The same expressions hold for N(p,Sk) as well. 4 JAITRA CHATTOPADHYAY, BIDISHA ROY, SUBHA SARKAR AND R. THANGADURAI

Proof. We prove for Q(p,Sk) and the proof of N(p,Sk) follows analogously. Let ψk be the characteristic function for Sk which is deﬁned as

1; if m Sk, ψk(m)= ∈ 0; if m Sk. 6∈ Now, by (2), we see that

p−1 p−1 1 m Q(p,S ) = f(m)= ψ (m)f(m)= ψ (m) 1+ k k 2 k p mX∈Sk mX=1 mX=1 (p−1)/k 1 p 1 1 k m (5) = − + , 2 k 2 p p mX=1 which proves (4). Then, by Theorem 4, we get 1 p 1 Q(p,S )= − + O(√p log p). k 2 k This finishes the proof. Let q > 1 be a positive integer and let ψ be a non-trivial quadratic character modulo ∞ ψ(n) q. Let L(s, ψ) = be the Dirichlet L-function associated to ψ. Since ψ is a ns Xn=1 non-trivial homomorphism, L(s, ψ) admits the following Euler product expansion ψ(p) L(s, ψ)= 1 − ps Yp∤q for all complex number s with (s) > 1. This, in particular, shows that L(s, ψ) > 0 for all real number s > 1. By continuity,ℜ it follows that L(1, ψ) 0. Dirichlet proved that L(1, ψ) = 0 in order to prove the infinitude of prime numbers≥ in an arithmetic progression. Hence,6 it follows that L(1, ψ) > 0 for all non-trivial quadratic character ψ. Since L(1, ψ) > 0, it is natural to expect some non-trivial lower bound as a function of q. This is what was proved by Landau-Siegel in the following theorem. The proof can be found in [14]. Theorem 5. Let q > 1 be a positive integer and ψ be a non-trivial quadratic character modulo q. Then for each ǫ> 0, there exists a constant C(ǫ) > 0 such that C(ǫ) L(1, ψ) > . qǫ The following lemma is crucial for our discussions. This lemma connects the sum of Legendre symbols and the Dirichlet L-function associated to Legendre symbol via the famous Dirichlet class number formula for the quadratic field. For an odd prime p, the Legendre symbol · = χ ( ) is a quadratic Dirichlet character modulo p. We also define p p · a character ( 1)(n−1)/2; if n is odd, χ4(n)= − 0; otherwise. DISTRIBUTION OF RESIDUES MODULO p USING THE DIRICHLET’S CLASS NUMBER FORMULA5

Then one can deﬁne the Dirichlet character χ4p as χ4p(n)= χ4(n)χp(n) for any odd prime p and similarly, we can deﬁne χ3p(n)= χ3(n)χp(n) for any odd prime p> 3. Clearly, χ4p and χ3p are non-trivial and real quadratic Dirichlet characters.

Lemma 2. (See for instance, Page 151, Theorem 7.2 and 7.4 in [24]) Let p > 3 be an odd prime and for any real number ℓ 1, we deﬁne ≥ (6) S(1,ℓ)= χp(m). 1≤Xm<ℓ Then we have the following equalities. (1) For a prime p 3 (mod 4), we have ≡ √p S(1,p/2) = (2 χ (2)) L(1, χ ), π − p p where L(1, χp) is the Dirichlet L-function; Also, we have √p S(1,p/3) = (3 χ (3))L(1, χ ). 2π − p p (2) For a prime p 1 (mod 4), we have ≡ √3p S(1,p/3) = L(1, χ ); 2π 3p Also, we have √p S(1,p/4) = L(1, χ ). π 4p Now, we need the following lemma, which deals with the vanishing sums of Legendre symbols. This was proved in [2]. For more such relations one may refer to [8]. Lemma 3. [2] Let p be an odd prime. Then the following equalities hold true. (p−1)/2 n (1) If p 1 (mod 4), then we have =0. ≡ p Xn=1 ⌊p/4⌋ n (2) If p 3 (mod 8), then we have =0. ≡ p Xn=1 ⌊p/2⌋ n (3) If p 7 (mod 8), then we have =0. ≡ p ⌈Xp/4⌉

3. Proof of Theorem 1

Let p be a given odd prime. We want to estimate the quantity Q(p,S2). Therefore, by (5), we get (p−1)/2 1 p 1 1 2 n (7) Q(p,S )= − + . 2 2 2 2 p p Xn=1 Now, we consider three cases as follows. Case 1. p 1 (mod 4) ≡ 6 JAITRA CHATTOPADHYAY, BIDISHA ROY, SUBHA SARKAR AND R. THANGADURAI

(p−1)/2 n In this case, since = 0, by Lemma 3 (1), the equation (7) reduces to p Xn=1 p 1 Q(p,S )= − , 2 4 which is as desired.

Case 2. p 3 (mod 8) ≡ By Lemma 2 (1) and by (7), we get 1 p 1 √p Q(p,S )= − + (2 χ (2)) L(1, χ ). 2 2 2 π − p p

2 In this case, we know that p = 1. Therefore, we get − 1 p 1 √p Q(p,S )= − +3 L(1, χ ). 2 2 2 π p 1 Let ǫ be any real number such that 0 <ǫ< 2 . Then by Theorem 5, we get

1 p 1 1 Q(p,S ) − p 2 −ǫ, 2 − 2 2 ≫ǫ as desired.

Case 3. p 7 (mod 8). ≡ 2 Since p 7 (mod 8), we know that = 1. Therefore, by Lemma 2 (1) and by (7), ≡ p we get 1 p 1 √p 1 p 1 √pL(1, χp) Q(p,S )= − + L(1, χ )= − + . 2 2 2 π p 2 2 π 1 Let ǫ be any real number such that 0 <ǫ< 2 . Then by Theorem 5 we get

1 p 1 1 Q(p,S ) − p 2 −ǫ 2 − 2 2 ≫ǫ which proves the theorem.

4. Proof of Theorem 2

Let p be a given odd prime. We want to estimate the quantity Q(p,S3). Therefore, by (5), we get,

(p−1)/3 1 p 1 3 n (8) Q(p,S )= − + . 3 2 3 p p Xn=1 Now, we consider the following cases.

Case 1. p 1 (mod 12) ≡ DISTRIBUTION OF RESIDUES MODULO p USING THE DIRICHLET’S CLASS NUMBER FORMULA7

3 Note that, in this case, we have = 1. By (8) and by Lemma 2 (2), we get p 1 p 1 1 √3p Q(p,S ) − = L(1, χ χ ) 3 − 2 3 2 2π 3 p √3p C(ǫ) ≥ 4π (3p)ǫ 1 p 2 −ǫ, ≫ǫ 1 for any given 0 <ǫ< 2 in Theorem 5. Case 2. p 11 (mod 12) ≡ 3 In this case, we have, = 1. Then again by (8) and by Lemma 2 (1), we get p 1 p 1 1 √3p Q(p,S ) = − + (3 χ (3))L(1, χ ). 3 2 3 2 2π − p p Hence 1 p 1 1 Q(p,S ) − p 2 −ǫ, 3 − 2 3 ≫ǫ 1 for any 0 <ǫ< 2 in Theorem 5.

5. Proof of theorem 3 At ﬁrst, using the equation (5), we note that (p−1)/4 (p−1)/4 1 p 1 1 4 m 1 p 1 1 m (9) Q(p,S )= − + = − + . 4 2 4 2 p p 2 4 2 p mX=1 mX=1

Case 1. p 1 (mod 4) ≡ Now, we apply Lemma 2 (2) in (9) and we get 1 p 1 1 √p Q(p,S ) = − + L(1, χ χ ). 4 2 4 2 π 4 p Hence p 1 1 Q(p,S ) − p 2 −ǫ, 4 − 8 ≫ǫ 1 for any 0 <ǫ< 2 in Theorem 5. Case 2. p 3 (mod 8) ≡ [(p−1)/4] m In this case, we apply Lemma 3 (2) which says that = 0. Hence, by (9), p Xn=1 we get 1 p 1 Q(p,S )= − . 4 2 4 Case 3. p 7 (mod 8) ≡ 8 JAITRA CHATTOPADHYAY, BIDISHA ROY, SUBHA SARKAR AND R. THANGADURAI

First note that by Lemma 3 (3), we have m =0. p−1 p−1 p 4

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(Jaitra Chattopadhyay, Bidisha Roy, Subha Sarkar and R. Thangadurai) Harish-Chandra Re- search Institute, HBNI, Chhatnag Road, Jhunsi, Allahabad 211019, India E-mail address, Jaitra Chattopadhyay: [email protected] E-mail address, Bidisha Roy: [email protected] E-mail address, Subha Sarkar: [email protected] E-mail address, R. Thangadurai: [email protected]