cp's OEIS Frontend

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).

%I A320060 #43 Mar 11 2023 06:56:46
%S A320060 -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,
%T A320060 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,
%U A320060 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
%N 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).
%C A320060 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).
%C A320060 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).
%C A320060 These two conjectures were motivated by Theorem 1.2(ii) in the author's preprint arXiv:1809.07766.
%C A320060 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).
%C A320060 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
%H A320060 Zhi-Wei Sun, <a href="/A320060/b320060.txt">Table of n, a(n) for n = 2..400</a>
%H A320060 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1809.07766">Quadratic residues and related permutations and identities</a>, arXiv:1809.07766 [math.NT], 2018.
%H A320060 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1810.12102">On quadratic residues and quartic residues modulo primes</a>, arXiv:1810.12102 [math.NT], 2018.
%e A320060 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).
%t A320060 rMod[m_,n_]:=rMod[m,n]=Mod[m,n,-n/2];
%t A320060 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];
%t A320060 Table[a[Prime[n]],{n,2,80}]
%Y A320060 Cf. A000040, A000290, A319311.
%K A320060 sign
%O A320060 2,5
%A A320060 _Zhi-Wei Sun_, Oct 18 2018