TITLE: Complex roots of an irreducible polynomial
QUESTION [5 upvotes]: This is a part of a question from a Berkeley prelim exam and I would appreciate a hint, since I can't see a promising approach to this.
Let $p$ be an irreducible polynomial over the rationals with a nonzero complex root $a$. Suppose that $a^2$ is also a root of $p$. How would one conclude from this that for any root $b$ of $p$, $b^2$ is also a root of $p$?

REPLY [3 votes]: Hint: The number $a$ is a zero of the polynomials $p(x)$ and $q(x)=p(x^2)$. Show that this means that $a$ is also a root of the greatest common divisor $h(x)=gcd(p(x),q(x))\in\mathbf{Q}[x]$. What can you say about the polynomial $h(x)$ using the fact that $p(x)$ was known to be irreducible?