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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?

Maestro13
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  • If one prime factor of $x^2 + 4$ is congruent to $3 \bmod 4$, then this must also be true of another prime factor, since $3^2 \equiv 1 \bmod 8$. – Mr. Brooks Feb 14 '17 at 22:32
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    Yes obviously. In fact one of the factors will be $\equiv 3 \mod 8$ and another then has to be $\equiv 7 \mod 8$. But that does not tell us whether it can occur at all. – Maestro13 Feb 14 '17 at 22:42
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    This may or may not be of any help: given $p \equiv 3 \bmod 8$ and $q \equiv 7 \bmod 8$, the Kronecker symbol $$\left(\frac{pq}{4}\right) = -1.$$ – Mr. Brooks Feb 14 '17 at 22:50

2 Answers2

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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.

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

kub0x
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