Can a factor of $x^2+4$ for odd $x$ be $\equiv 3\mod 4$? When $x$ is odd, then $x^2+4 \equiv 5 \mod 8$. Can any factor of $x^2+4$ be $\equiv 3\mod 4$?
Examples:
$$7^2+4=53$$
$$11^2+4= 5^3$$
$$31^2+4= 5 \cdot 193$$
$$47^2+4=2213$$
$$89^2+4=5^2 \cdot 317$$
All of the prime factors above are $\equiv 1 \mod 4$.
Is there no counter example and if so, how to prove this?

If $p$ is an odd prime divisor of $x^2+4$, then we have $-4 \equiv x^2 \pmod p$
So $1=\left(\frac{-4}{p}\right)=\left(\frac{4}{p}\right) \left(\frac{-1}{p}\right) = \left(\frac{-1}{p}\right)$ which means that $p \equiv 1 \pmod 4$ by the first supplement to quadratic reciprocity.
