Let $f\in\mathbb{Z}[X]$ be a monic polynomial with $\text{deg}(f)=5$. Suppose that there exist a prime number $p$ and a finite field $F$ of order $p^2$ such that $f$ has no roots in $F$. Prove that $f$ is irreducible in $\mathbb{Z}[X]$.
I've been trying to prove this practice problem for a while now but I can't seem to really understand how to do it. If anyone can help me with this problem, it would be much appreciated.
Edit: This was just a practice problem from an old exam which had no answers. I really wasn't able to do that much of it myself. I went on on the wrong direction for a long time. Thanks for the help!
Let's do the contrapositive. Assume $f$ is reducible over $\mathbb{Z}$. If it has a root in $\mathbb{Z}$, then it has a root in $\mathbb{F}_p$ for every prime $p$, and therefore it has a root in every field of order $p^2$ (which contains the corresponding field of order $p$).
If $f$ is reducible over $\mathbb{Z}$ but has no roots, then it must be a product of an irreducible quadratic and an irreducible cubic (since $f$ has degree $5$). If $p$ is a prime, then the reduction of $f$ will also be expressible as a product of a quadratic and a cubic; if either of them is reducible modulo $p$, then it has a root in $\mathbb{F}_p$, and therefore in the corresponding field of $p^2$ elements. Otherwise, both the quadratic and the cubic are irreducible. But an irreducible quadratic over $\mathbb{F}_p$ always has a root in the field of $p^2$ elements, so in this case, $f$ will also have a root in every field of order $p^2$.
Thus, if $f$ is reducible over $\mathbb{Z}$, then it has a root in every field of $p^2$ elements. By contrapositive, if there is a field with $p^2$ elements in which $f$ has no roots, then it must be irreducible over $\mathbb{Z}$.
The claim also holds for degrees less than $5$, but not for degree $6$, since the polynomial could factor into a product of two irreducible cubics, neither of which has roots modulo $p$ (and therefore neither has roots in the field with $p^2$ elements).