Irreducible polynomial with $\alpha^l=\beta^l \in \mathbb{Q}$ for any two roots

Question

Let $f$ be an irreducible polynomial in $\mathbb{Q}[x]$ with $\alpha^l=\beta^l \in \mathbb{Q}$ for any two roots $\alpha,\beta \in \mathbb{C}$ of $f$. Can I follow $f | (x^l-\alpha^l)$ from this? If $\deg f \leq l$, the statement is obvious. But is $\deg f > l$ possible?

Answer 1

Let $c=\alpha^l$ and $g(X)=X^l-c$. As $f$ is irreducible, $\gcd(f,g)$ is either $1$ or $f$ itself. In the latter case, of course $f\mid g$ and $\deg f\le \deg g=l$. In the first case, $f$ has no root in common with $g$, whereas all roots of $f$ must be roots of $g$; then $f$ has no roots at all and is constant.