TITLE: Prove that $x^2+1$ is reducible in $\mathbb{Z}_p[x]$.
QUESTION [5 upvotes]: Prove that $x^2+1$ is reducible in $\mathbb{Z}_p[x]$ if and only if there exists integers $a$ and $b$ such that $a+b=p$ and $ab \equiv 1 \pmod{p}$. (Here $\mathbb{Z}_p$ means the integers modulo a prime $p$).
I'm having trouble with the 'only if' direction of this proof, the 'if' direction is pretty trivial. After assuming $f$ is reducible, I've been able to use the Factor Theorem to show that $f$ is of the form $f(x)=c(x-a)(x-b)$ for $a,b,c \in \mathbb{Z}_p$, so $x^2+1=cx^2-c(a+b)x+c(ab)$. Comparing coefficients, we see that $c=1$, so we must have $ab=1$ and $a+b=0$, hence, $a+b \equiv 0 \pmod{p}$ and $ab \equiv 1 \pmod {p}$. Id like to show that $a+b=p$. This seems to be the case, but I'm having difficulty proving it.
Is it okay here to assume that $a<p$ and $b<p$ (since they are elements of $\mathbb{Z}_p$), and assume without loss of generality that $a\leq b$, so that $a\leq b < p$, and use this to deduce that if $a+b=\alpha p$ for some $\alpha \in \mathbb{Z}$ with $\alpha \geq 0$, then we must have $\alpha =1$? 
I want to say something like, if either $a=0$ or $b=0$, then $ab \equiv 0 \pmod{p}$, a contradiction, so we must have $a+b\geq 2$, so we cannot have $\alpha=0$. If $2\leq \alpha$, then since $a\leq b < p$, we have $a+b<p+b<2p\leq \alpha p$, contradicting that fact that $a+b=\alpha p$, so we must have $\alpha =1$. I have a strong feeling I'm over-thinking this problem, thanks in advance for any help!

REPLY [3 votes]: Yes: The "largest" $a$ and $b$ can be is $p-1$, so the largest $a+b$ could be is $2p-2$. This shows $\alpha=1$.