Prime factors of $16k^4 +1$ mod $8$

Question

I need to show that for $k \in \mathbb{Z}$ there exists no prime factor $p$ of $16k^4 + 1$ with $p \equiv -1 \pmod 8$. How would I approach this problem?

Answer 1

If $16k^4 + 1 \equiv 0 \bmod p$, then $(2k)^4 \equiv -1 \bmod p$ and so $2k$ has multiplicative order $8$ mod $p$. By Lagrange's theorem, $8$ must divide $p-1$, that is $p \equiv 1 \bmod 8$.