Showing $p \mid 16x^4 + 1 \implies p \equiv 1 \pmod 4$

Question

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!

Answer 1

If $p|16x^4+1$, we can write that as $16x^4+1\equiv0\pmod{p}$, or

$$16x^4\equiv-1\pmod{p}$$

Since $16x^4=(4x^2)^2$ is a perfect square, we have that $-1$ is a quadratic residue, mod $p$, which is true if and only if $p\equiv1\pmod{4}$.

Answer 2

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 $p\equiv 1\mod 4$.

Answer 3

We are going to do the case-by-case type of solution and find the remainder of $p$ mod $4$. Obviously the case $p$ is even is excluded as the RHS is an odd number. So we are left to show that $p$ can't be congruent to $3$ mod $4$. But this is actually kind of a famous results. The steps are:

1) If $p$ is of the form $4k+3$, then $p$ would have a divisor of the same form - call this prime divisor $q$.

2) $q|16x^4+1$ means $q|(4x)^2+1 \implies -1 \equiv (4x)^2\pmod{q}$. Raise this to the power of $\frac{q-1}{2}$, which is odd as $q\equiv3\pmod{4}$, therefore we end up with $-1 \equiv (4x)^{q-1} \pmod{q}$ But this is a contradiction with Fermat's Theorem which in this case says that

$(4x)^{q-1} \equiv 1\pmod{q}$ since $\gcd(q, 4x)=1$

Hopefully this helped at least a bit.

Answer 4

Indeed more is true. If $p\mid 16x^4+1$ then $p$ is odd, and $(2x)^4\equiv-1\pmod p$. This means that $2x$ has multiplicative order $8$ modulo $p$. That implies $8\mid(p-1)$, that is $p\equiv1\pmod8$.