Let $\mathbb{F}_{p}$ be the field with $p$ elements. Determine all primes $p$ for which $x^{2}+1$ is irreducible over $\mathbb{F}_{p}$.
Here's what I've got. I think this might be finished but I'm not sure.
If $p=2$ then 1 is a root of $x^{2}+1$ so is reducible over $\mathbb{F}_{p}$. So assume $p>2$. We have $x^{2}+1$ is reducible $\iff$ it has a root $\alpha$ in $\mathbb{F}_{p} \iff \alpha^{2}=-1 \implies \alpha^{4}=1$. Since $\alpha^{3}=-\alpha$ and $-\alpha \neq 1$ when $p>2$ we have $|\alpha| = 4$ in $\mathbb{F}^{\times}_{p}$. Therefore $4$ divides $p-1$ by Lagrange, i.e. $p\equiv 1$ mod $4$ therefore $x^{2}+1$ is irreducible for all primes $p>2$ not congruent to $1$ mod $4$.
For the converse, evidently this works:
Since $p$ congruent to $1$ mod $4$ we have by Fermat that $p$ is the sum of squares, i.e. $p=a^{2}+b^{2}$. So if $c=(ab)^{-1}$ in $\mathbb{F}_{p}$ then $x^{2}+1=(x+ca^{2})(x+cb^{2})$ so is reducible.
Nope, you're not missing anything; all of the $\implies$ you wrote down are really $\iff$.
To be more explicit about the other direction: if $p\equiv 3\bmod 4$, then $\mathbb{F}_p^\times$ is a cyclic group of order $p-1\equiv 2\bmod 4$, so that there cannot be an element $\alpha\in\mathbb{F}_p^\times$ of order $4$ by Lagrange's theorem, so that the polynomial $x^2+1$ cannot have roots in $\mathbb{F}_p$, so that the polynomial $x^2+1$ is irreducible over $\mathbb{F}_p$.