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A227968
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Determinant of the (p_n+1)/2 X (p_n+1)/2 matrix with (i,j)-entry (i,j=0,...,(p_n-1)/2) equal to the Legendre symbol((i^2+j^2)/p_n), where p_n is the n-th prime.
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8
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-1, 2, -12, -80, -162, 3528, -9216, -11264, 482230, -206684160, 1488942450, 976835722500, -1420648513536, -12993312063488, -4001622478404278, -738964000238206976, 73685520670239843750, -13192753286712605540352, -505689449431040
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OFFSET
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2,2
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COMMENTS
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Conjecture: 2*a(n) is always a quadratic residue mod p_n.
Zhi-Wei Sun also made the following general conjecture:
Let p be any odd prime. For each integer d let T(d,p) be the determinant of the (p+1)/2 X (p+1)/2 matrix whose (i,j)-entry (i,j=0,...,(p-1)/2) is the Legendre symbol ((i^2+d*j^2)/p). Then T(-d,p) == (2/p)*T(d,p) (mod p). If d is a quadratic residue mod p, then so is 2*T(d,p). If d is a quadratic non-residue mod p, then T(d,p) is a quadratic residue mod p.
The author has proved the following results for any odd prime p:
(1) If c and d are integers with c not divisible by p, and d' == c^2*d (mod p), then T(d',p) = (c/p)^((p+1)/2)*T(d,p).
(2) If p is congruent to 1 mod 4 and d is not divisible by p, then T(-d,p) = (2/p)*T(d,p).
The general conjecture mentioned above was proved in a recent preprint of Zhi-Wei Sun.
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LINKS
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EXAMPLE
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a(2) = -1 since the determinant |((i^2 + j^2)/3)|_{i=0,1; j=0,1} equals -1.
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MATHEMATICA
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a[n_] := Det[Table[JacobiSymbol[i^2+j^2, Prime[n]], {i, 0, (Prime[n]-1)/2}, {j, 0, (Prime[n]-1)/2}]]; Table[a[n], {n, 2, 20}]
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PROG
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(PARI) a(n) = my(p=prime(n)); matdet(matrix((p+1)/2, (p+1)/2, i, j, i--; j--; kronecker(i^2+j^2, p))); \\ Michel Marcus, Aug 25 2021
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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