To show that a polynomial (over a field) of form $p(x^2)$ is irreducible, i was suggested to show that $p$ is irreducible and conclude from that that $p$ would be irreducible. For example take $q = x^4-16x^2+4$. Then $q = p(x^2)$ where $p=x^2-16x+4$. It is clear that $p$ is irreducible. However i can't see how this would imply that $q$ is also irreducible.
For example $x^4-x^2 = (x^2+x)(x^2-x)$ factors in a way that has non even powers of x in it. However $x^2-x$ still happens to be reducible in this case so it's not counterexample...