Can a factor of $x^2+4$ for odd $x$ be $\equiv 3\mod 4$?

Question

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?

Answer 1

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.

Answer 2

$x^2 + 4 \equiv 3 \pmod 4$ 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 $4n(n+1) + 1\equiv 1 \pmod 4$