<s>[INST] 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?
 [/INST] $x^2 + 4 \equiv 3 \pmod 4$ is the same as proving $x^2 + 0 \equiv 3 \pmod 4$ since $4 \equiv 0\pmod 4$.
If $x$ is even is easy to see that $x^2 \equiv 0 \pmod 4$ and when $x$ is odd it has the form $2n+1$ then:
$x=(2n+1)^2= 4n^2 + 4n + 1 = 4n(n+1) + 1$ thus $4n(n+1) \equiv 0 \pmod 4$ and $4n(n+1) + 1\equiv 1 \pmod 4$
</s>