$17$ is a quadratic residue for all primes $p$ such that $p \equiv \pm 3 \mod{8}$?

Question

Trying to prove that $17$ is a quadratic residue for all primes $p$ such that $p \equiv \pm 3 \mod{8}$?

Thanks!

Answer 1

As Daniel points out the statement is wrong, indeed there are infinitely many primes in both progressions for which 17 is a quadratic non-residue and infinitely many primes for which it is a quadratic residue. To see why, using quadratic reciprocity, 17 is a quadratic residue mod $p$ if and only if $p$ is a quadratic residue mod 17 that is if $p\equiv, 1,2,4,8,9,13,15,16 \pmod{17}$.

Now look for example the primes in the following arithmetic progressions

• all primes $p=8\cdot 17k+3$: are in the progression $8k+3$ and are $\equiv 3 \pmod{17}$ and so they are quadratic nonresidues mod 17. Likewise all primes $p=8\cdot 17k+19$, are $\equiv 2 \pmod{17}$ and so are quadratic residues mod 17.

• In the same way primes $p=8\cdot 17k+5$ are quadratic non-residues mod 17 and primes $p=8\cdot 17k+21$, are all quadratic residues mod 17, and they are all in the progression $8k-3$.

Dirichlet theorem impies that there are infinitely many primes in each of this progressions.