I am trying to solve a homework exercise in elementary number theory:     Which primes $p$ satisfy $n^2 \equiv -1 \mod p$ for a perfect square $n^2$?    After looking at the case $p=5$, I saw that $3^2 \equiv 4 \mod 5$, but $p=2^2+1$, I thought that maybe the answer would be primes $p$ such that $p=m^2+1$ for some $m$. Certainly this would imply that there is an $n$ so that $p \mid n^2+1$ (in particular $n=m$).  Unfortunately, $n=13$ doesn't satisfy that condition. However, weakening it to $p\mid n^2+1$ for some $n$ is just the statement of the problem. I don't want to answer "the congruence is true for primes for which it is true." So I am back to square one.
Let $p$ be a prime number. If -1 is a square in $\mathbf F_p$, it means that there exists an element of order 4. So $p-1$ is divisible by 4 and therefore $p \equiv 1$ (mod 4). I wonder the converse is also true. Does $p \equiv 1$ (mod 4) mean the existence of an element whose square is $-1$ in $\mathbf F_p$?
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