Let $p$ be an odd prime. Show that there exists integers $x$ and $y$ with $x^2 + 1 = py$ and if and only if $p \equiv 1 \pmod 4$.

Question

Let $p$ be an odd prime. Show that there exists integers $x$ and $y$ with $x^2 + 1 = py$ and if and only if $p \equiv 1 \pmod 4$.

Don't know where to start. Please help! All I know is that Wilson's theorem and Fermat's could possibly help. Don't know what the $y$ does/how it relates to the problem.

@gnasher729 $p$ is assumed to be odd. Otherwise this is indeed false, as you point out. — AnCar, Nov 19 '23 at 10:06
The duplicate uses Wilson and little Fermat, which you were supposed to use here, as you indicated. Next time you should try to search for it before saying "Don't know". — Dietrich Burde, Nov 19 '23 at 11:29

score 1 · Answer 1 · answered Nov 19 '23 at 10:05

This is an application of Euler's criterion. If $p$ is an odd prime and $a$ is a number coprime with $p$, then \begin{equation} a^{\tfrac{p-1}{2}}\equiv \begin{cases} 1 & \quad \text{if there exists an integer } x \text{ such that } x^2\equiv a \mod p \\ -1 & \quad \text{if there is no such integer} \end{cases} \end{equation}

Two elementary proofs of this can be found on wiki: https://en.wikipedia.org/wiki/Euler%27s_criterion

Let us apply this with $a=p-1$, which is obviously coprime with $p$.

If $p=4k+1$, then $(p-1)^{\tfrac{p-1}{2}}=(p-1)^{2k}\equiv (-1)^{2k}\equiv 1\mod p$, so by Euler's criterion, there exists some integer $x$ such that $x^2\equiv p-1\mod p$, meaning that $x^2+1=py$ for some integer $y$.

If $p=4k+3$, then $(p-1)^{\tfrac{p-1}{2}}=(p-1)^{2k+1} \equiv (-1)^{2k+1} \equiv -1\mod p$, so by Euler's criterion, there can exist no integer $x$ such that $x^2+1\equiv 0\mod p$.

Let $p$ be an odd prime. Show that there exists integers $x$ and $y$ with $x^2 + 1 = py$ and if and only if $p \equiv 1 \pmod 4$.

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