I know that in a field $F$ with character $p$($p$ is a prime number), we have $x^{p^n}-x-a \,(a \in F)$ is reducible \iff $x^{p^n}-x-a$ has root in $F$ Then the question is to prove $x^{p^n}-x-a \, \equiv \, 0 \, $(mod $p$) has solution.
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Very closely related. – Jyrki Lahtonen Dec 17 '19 at 16:24
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I don't think that polynomial has a solutions in $GF(p)$ (unless $a=0$), but it is reducible. If $\alpha$ is a root (in some extension field), then $\alpha^{p^n}=\alpha+a$ implying that $\alpha^{p^{np}}=\alpha$. Therefore the number of conjugates of $\alpha$ over $GF(p)$ is at most $np$ which is less than $p^n$, and we are done. – Jyrki Lahtonen Dec 17 '19 at 16:28
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I closed this as a duplicate assuming that the formulation in your title is the intended one. If not, please clarify. – Jyrki Lahtonen Dec 17 '19 at 16:29
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Thank you . That equivalence is not true. – user725757 Dec 18 '19 at 01:05