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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 $p$?

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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?

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    Or in another way: $p$ is the minimal polynomial of $a$ over $\mathbb{Q}$. Hence we have $\mathbb{Q}(a) \cong \mathbb{Q}\left\[x \right\] /(p) \cong \mathbb{Q}(b)$ for every root $b$ of $p$ via $a \mapsto \overline{x} \mapsto b$ and since $a^2$ is$a$root of $p$ we get $\overline{x}^2$ is$a$root of $p$ and therefore $b^2$ is$a$root of $p$ for every root $b$ of $p$.2011-11-03