Show $p(X)$ (over a field) is irreducible iff $p(X+a)$ is irreducible
Solution 1:
One way to prove this is to instead prove that $p(X)$ is reducible if and only if $p(X + a)$ is reducible.
To this end, note that if $p(X) = q(X)r(X)$, then $p(X+a) = \hat{q}(X)\hat{r}(X)$ where $\hat{q}(X) = q(X + a)$ and $\hat{r}(X) = r(X + a)$. Going the other way, if $p(X+a) = q(X)r(X)$, then $p(X) = \tilde{q}(X)\tilde{r}(X)$ where $\tilde{q}(X) = q(X - a)$ and $\tilde{r}(X) = r(X - a)$.
Solution 2:
Hint: Show that $\phi_a:A[X]\to A[X]$ defined as $p(X)\mapsto p(X+a)$ is an automorphism.