OFFSET
2,2
COMMENTS
Conjecture 1: Let n be any positive integer.
(i) If a(2*n) is nonzero, then 4*n + 1 is a sum of two squares.
(ii) a(2*n + 1) is divisible by phi(4*n + 3)/2, where phi is Euler's totient function. If n is even, then a(2*n + 1)/(phi(4*n + 3)/2) is a square. This has been verified for n = 2..1000.
For any odd integer n > 3 and integers c and d, we introduce the notation: {c,d}_n = det[Jacobi(i^2 + c*i*j + d*j^2, n)]_{1 < i, j < n-1}.
The following conjecture is similar to Conjecture 1.
Conjecture 2: (1) {2, 2}_p = 0 for any prime p == 13,19 (mod 24), and {2, 2}_p == 0 (mod p) for any prime p == 17,23 (mod 24).
(2) If n == 5 (mod 8), then {4, 2}_n = 0. If n == 5 (mod 12), then {3, 3}_n = 0.
(3) If n == 5 (mod 12) and n is a sum of two squares, then {10, 9}_n = 0. Also, {10, 9}_p == 0 (mod p) for any prime p == 11 (mod 12).
(4) {8, 18}_p == 0 (mod p^2) for any prime p == 19 (mod 24), and {8,18}_p == 0 (mod p) for any prime p == 23 (mod 24). If n == 13,17 (mod 24) and n is a sum of two squares, then {8, 18}_n = 0.
We have verified Conjecture 2 for p or n smaller than 2000.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 2..73
D. Krachun, F. Petrov, Z.-W. Sun and M. Vsemirnov, On some determinants involving Jacobi symbols, Finite Fields Appl. 64 (2010), Article 101672.
Z.-W. Sun, On some determinants with Legendre symbol entries, Finite Fields Appl. 56 (2019), 285-307.
EXAMPLE
a(2) = 1 since the determinant of the matrix [Jacobi(i^2 + 3*i*j + 2*j^2, 5)]_{1 < i, j < 2*2} = [1,0; 0,1] is 1.
MATHEMATICA
a[n_]:=a[n]=Det[Table[JacobiSymbol[i^2+3*i*j+2*j^2, 2n+1], {i, 2, 2n-1}, {j, 2, 2n-1}]];
tab={}; Do[tab=Append[tab, a[n]], {n, 2, 29}]; Print[tab]
PROG
(PARI) f(i, j) = i^2 + 3*i*j + 2*j^2;
a(n) = matdet(matrix(2*n-2, 2*n-2, i, j, kronecker(f(i+1, j+1), 2*n+1)));
vector(25, n, a(n+1)) \\ Michel Marcus, Apr 27 2024
CROSSREFS
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Apr 27 2024
STATUS
approved