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Let $q=p^n$, $p$ prime, ¿Which $q$ satisfies $\mathbb{F}_{q^2}=\mathbb{F}_q(\sqrt{a})$?. I don't know why is necessary a condition for $q$.

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See the answer to this related question for ideas – Dilip Sarwate Sep 30 '12 at 23:30
up vote 1 down vote accepted

Hint: If $p\ne 2$, any polynomial $x^2+ax+b$ can be written in the form...

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