Show $8K^2-1$ has atleast one prime factor congruent to $-1mod8$

Question

Given that every odd prime factor of $n^2-2$ is congruent to $1 \mod 8$ or $-1 \mod 8$, how do I deduce from this a number of the form $8k^2-1$, $k$ an integer, must have at least one prime factor that is congruent to $-1 \mod 8$?

Answer 1

Let $p$ be a prime dividing $8k^2-1$. Then of course $2$ is a square $\pmod p$ which implies (as you point out) that $p\equiv \pm 1 \pmod 8$. But if all such $p$ were $1\pmod 8$ then so would $8k^2-1$ be, but that is clearly false.