Suppose that $f(x) = x^4 + a \in \mathbb{Z}[x]$ is reducible over $\mathbb{Q}$ with $|a| > 1$, how can I see that then $|a|$ must be a square in $\mathbb{Z}$?
Not sure if there is any straightforward way here?
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Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you [edit] your question to include some motivation, and an explanation of your own attempts. – Christian E. Ramirez Sep 27 '22 at 14:39
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It also follows from this post. – Dietrich Burde Sep 27 '22 at 14:42
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@DietrichBurde I don't see why it follows from that post. Could you explain? – LocationMap2 Sep 27 '22 at 15:00
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$x^4+ax^2+b$ reduces to $x^4+b$ for $a=0$. Then the post just says the same. So $x^4+b$ is irreducible if and only if the square of the roots, etc. is not in $\Bbb Q$. But this just amounts to saying that $-b$ is not a square. – Dietrich Burde Sep 27 '22 at 15:18