Why is $\sqrt{5}$ an element of every field of order $p^{2 e}$?

Question

This was claimed in an answer to another question I asked but it's unclear to me why it's true. I'd also be happy with a reference that explains it. Thanks!

Answer 1

Let's see that $x^2-5$ has a root in every field of order $p^2$. If it's irreducible over the field $F$ with $p$ elements, then it has a root in the field $F[x]/(x^2-5)$ that has $p^2$ elements. Since, up to isomorphism, there is only one field of $p^2$ elements, we're done.

If $F$ is a field with $p^{2e}$ elements, then it has a subfield with $p^2$ elements, because $2$ divides $2e$.