# Irreducible polynomial mod p^2

Consider the polynomial $x^{2} + p$ modulo $p^{2}$.
Suppose $x^2+p\pmod{p^2}$ is reducible. Then, $x^2+p\equiv 0\pmod{p^2}$ has a solution. By reducing the modulus, we have $x^2\equiv 0\pmod{p}$, so $x$ is divisible by $p$. Write $x=py$. Putting this back into the original equation gives $(py)^2+p\equiv p^2y^2+p\equiv p \equiv 0\pmod{p^2}$, which is certainly not true.
Therefore, $x^2+p\pmod{p^2}$ is irreducible.