A320060 Least residue greater than -prime(n)/2 of the product Product_{i,j} (i^2 - (i+j)*j) modulo prime(n), where i and j run over {1,...,(prime(n)-1)/2} with i^2-(i+j)*j not divisible by prime(n).

-1, -1, -1, 1, 5, -4, -1, 1, 1, 1, -6, 1, -1, -1, -23, -1, 1, 1, 1, 27, -1, -1, 1, 22, -1, 1, 1, -1, 15, -1, 1, 37, -1, -1, 1, 28, -1, -1, 80, -1, -1, 1, -81, 14, -1, 1, 1, 1, 1, -89, -1, 1, 1, 16, 1, -1, 1, 60, 1, -1, -138, 1, 1, -25, -114, 1, 148, 1, 1, -42, -1, -1, -104, -1, 1, -1, 63, -1, -1

Offset: 2

Author

Zhi-Wei Sun, Oct 18 2018

Keywords

• sign

Comments

Conjecture 1. Let p be an odd prime and let f(p) be the product of i^2-(i+j)*j, where i and j run over {1,...,(p-1)/2} with i^2-(i+j)*j not divisible by p. If p == 1, 9 (mod 20), then f(p) == -5^((p-1)/4) (mod p). If p == 11 (mod 20) or p == 23, 27 (mod 40), then f(p) == 1 (mod p). If p == 19 (mod 20) or p == 3, 7 (mod 40), then f(p) == -1 (mod p). If p == 13 (mod 20) then f(p) == (-1)^(floor((p+10)/20))*5^((p-1)/4) (mod p); if p == 17 (mod 20) then f(p) == (-1)^(floor((p-10)/20))*5^((p-1)/4) (mod p).

Conjecture 2. Let p be any prime congruent to 1 modulo 4, and let a and b be integers with a*b == -1 (mod p). Let F(a,b) be the product of (i-a*j)*(i-b*j), where i and j run over {1,...,(p-1)/2} with (i-a*j)(i-b*j) not divisible by p. If a-b is not divisible by p, then -F(a,b) is congruent to the Legendre symbol ((a-b)/p) modulo p. If a == b (mod p) and p == 1 (mod 8), then F(a,b) == (-1)^((p+7)/8)*((p-1)/2)! (mod p). If p == 5 (mod 8) and a == b == (-1)^k*((p-1)/2)! (mod p) with k in {0,1}, then F(a,b) == (-1)^(k+(p-5)/8) (mod p).

These two conjectures were motivated by Theorem 1.2(ii) in the author's preprint arXiv:1809.07766.

The author has proved Conjecture 2 fully and Conjecture 1 in the case p == 1,9 (mod 10) in arXiv:1810.12102. Here we add a new general conjecture which implies Conjecture 1 in the case p == 3,7 (mod 10).

Conjecture 3. Let A be an integer, and let p be an odd prime with A^2+4 a quadratic nonresidue modulo p. Let f(p,A) be the product Product_{i,j=1,...,(p-1)/2} (i^2-Aij-j^2). If p == 1 (mod 4), then f(p,A) == (-A^2-4)^((p-1)/4) (mod p). If p == 3 (mod 4) then f(p,A) == (-A^2-4)^((p+1)/4)*u_{(p+1)/2}(A)/2 (mod p), where u_0(A) = 0, u_1(A) = 1, and u_{n+1}(A) = A*u_n(A) + u_{n-1}(A) for n = 1,2,3,.... - Zhi-Wei Sun, Oct 30 2018

Examples

			a(4) = -1 since prime(4) = 7 does not divide i^2-(i+j)*j for any i,j = 1,2,3, and Product_{i,j = 1,2,3} (i^2 - (i+j)*j) = -108900 == -1 (mod 7).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 2..400
• Zhi-Wei Sun, Quadratic residues and related permutations and identities, arXiv:1809.07766 [math.NT], 2018.
• Zhi-Wei Sun, On quadratic residues and quartic residues modulo primes, arXiv:1810.12102 [math.NT], 2018.

Crossrefs

Cf. A000040, A000290, A319311.

Programs

Mathematica

rMod[m_,n_]:=rMod[m,n]=Mod[m,n,-n/2];
a[p_]:=a[p]=rMod[Product[If[rMod[i^2-(i+j)*j,p]==0,1,i^2-(i+j)*j],{i,1,(p-1)/2},{j,1,(p-1)/2}],p];
Table[a[Prime[n]],{n,2,80}]