A227609 Determinant of the (p_n-1)/2 X (p_n-1)/2 matrix with (i,j)-entry being the Legendre symbol((i^2+j^2)/p_n), where p_n is the n-th prime.

-1, 1, -4, -16, -27, 441, -1024, -1024, 34445, -13778944, 82719025, 48841786125, -67649929216, -564926611456, -153908556861703, -25481517249593344, 2456184022341328125, -399780402627654713344, -14448269983744, -214168150727821285287075

Offset: 2

Author

Zhi-Wei Sun, Jul 17 2013

Keywords

• sign

Comments

Conjecture: p_n never divides a(n), and moreover -a(n) is a quadratic residue mod p_n.

Zhi-Wei Sun also made the following conjecture:

Let p be any odd prime. For each integer d let S(d,p) be the determinant of the (p-1)/2 X (p-1)/2 matrix whose (i,j)-entry is the Legendre symbol ((i^2+d*j^2)/p). If d is a quadratic residue mod p, then so is -S(d,p). If d is a quadratic non-residue mod p, then we have S(d,p) = 0.

These were proved in version 9 of arXiv:1308.2900 (2018). In addition, the author has the following new conjecture.

Conjecture: For any prime p == 3 (mod 4), the number -S(1,p) is a positive square divisible by 2^((p-3)/2), i.e., -S(1,p) = (2^((p-3)/4)*m)^2 for some positive integer m. - Zhi-Wei Sun, Sep 09 2018

Examples

			a(2) = -1 since the Legendre symbol ((1^2 + 1^2)/3) is -1.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 2..100
• Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2019.
• Zhi-Wei Sun, Is -det[Legendre(i^2+j^2,p)]_{i,j=1,...,(p-1)/2} always a square for each prime p == 3 (mod 4)?, Question 310192 in MathOverflow, Sept. 9, 2018.

Crossrefs

Cf. A179071, A179072, A227968, A227971.

Programs

Maple

with(numtheory): with(LinearAlgebra):
a:= n-> Determinant(Matrix((ithprime(n)-1)/2, (i, j)->
        jacobi(i^2+j^2, ithprime(n)))):
seq(a(n), n=2..20);  # Alois P. Heinz, Jul 18 2013

Mathematica

a[n_]:=Det[Table[JacobiSymbol[i^2+j^2,Prime[n]],{i,1,(Prime[n]-1)/2},{j,1,(Prime[n]-1)/2}]]
Table[a[n],{n,2,20}]

PARI

a(n) = my(p=prime(n)); matdet(matrix((p-1)/2, (p-1)/2, i, j, kronecker(i^2+j^2, p))); \\ Michel Marcus, Aug 25 2021