cp's OEIS Frontend

From Jianing Song, Apr 21 2022: (Start)

Primes p such that kronecker(17, p) = kronecker(p, 17) = 1, where kronecker() is the kronecker symbol. That is to say, primes p that are quadratic residues modulo 17.

Primes p such that p^8 == 1 (mod 17).

Primes p == 1, 2, 4, 8, 9, 13, 15, 16 (mod 17). (End)

The primes are 3, 5 and those given in A191018 (Jacobi(prime|15) = +1).

For a(n) neither a multiple of 3 nor of 5 the Jacobi(a(n)|15) = +1.

The sequence (a(n)){n >=1} is the set S := {3^a*5^b*Product{j=1..m} (p_j)^{e(j)}}, in increasing order, with a and b from {0, 1}, primes p_j from A191018, m >= 0, and the exponents e(j) >= 0. If a = b = m = 0 then S = {1} and a(1) = 1.

The multiplicity of the (representative) solutions x is 2^m(n) for modulus a(n) from the set S. Thus it is 1 for 1, 3, 5, and a power of 2 with m(n) >= 1. This follows from Jacobi(prime|15) = +1, and the lifting theorem for powers of these primes (see e.g., Apostol). Primes 3 and 5 have only 1 solution and no lifting to powers >= 2 is possible. See A347833 for these multiplicities.

For the solutions x see A347832.

Originally incorrectly named "primes which are squares mod 86", which is sequence A106891. - M. F. Hasler, Jan 15 2016

This sequence is the complement of A033212 (primes congruent to 1 or 4 mod(15)) relative to the primes p with Jacobi(p|15) = +1 (A191018).

There is neither a solution x of the congruence x^2 == a(n) (mod 3) nor of x^2 == a(n) (mod 5) (the Legendre symbols are -1 in both cases, and Jacobi(a(n)|15) = +1).