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Let $F$ be a finite field and let $a,b \in F$. Prove that $x^2-a$, $x^2-b$ and $x^2-ab$ can't all be irreducible in $F[x]$ where $F$ is a finite field.

I tried to derive a contradiction by assuming $F(\sqrt{a}) = F(\sqrt{b}) = F(\sqrt{ab})$ where $[F(\sqrt{a}):F] = 2$ and then using theory about vector spaces (in particular, using that (1,$\sqrt{a}$) would be a basis), but could not find anything. Any suggestions?

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If $F$ is finite, $(F^*, \times)$ is cyclic generated by $u$, so $a$ is not a square implies $a=u^{2n+1}$, $b$ is not a square implies that $ b=u^{2m+1}$ implies $ab$ is a square.

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  • $\begingroup$ Nice one. Thanks. $\endgroup$
    – user2959071
    Dec 14, 2017 at 1:03
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    $\begingroup$ Another way of expressing this is $F/F^2\cong \mathbb{Z}_2$. $\endgroup$
    – Rene Schipperus
    Dec 14, 2017 at 1:25

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