Showing $p \mid 16x^4 + 1 \implies p \equiv 1 \pmod 4$ Essentially, I'm sure this is a simple answer as my lecturer skipped past this step. However I'm having trouble seeing why this is true. Is it making use of a theorem or is it simply taking everything $\pmod 4$ since $16$ will obviously be congruent to $0 \pmod 4$.
But even in this case, $p$ is a divisor so I don't see why that would be the case unless we stated $p = 16x^4 + 1$. Then it would follow that $p \equiv 1 \pmod 4$.
I wonder if it's something to do with Mersenne numbers but I feel like I am really missing the point and complicating something simple. 
Any help is appreciated!

In terms of congruences, it means $\;-1\equiv 16x^4=(4x^2)^2\mod p$. Thus it implies $-1$ is a square modulo $p$, and it is known this is true for an odd prime $p$ if and only if $p\equiv 1\mod 4$.
