Prove that $x^{p^{n}}-x-a$ is reducible over GF(p).

Question

Let $p$ is a prime,$n≥1$ and $n \nmid p^{n}$.Suppose $a\in GF(p)$.Prove that $x^{p^{n}}-x-a$ is reducible over $GF(p)$.

I just have no idea. Please help me,thanks!

Answer 1

$\DeclareMathOperator{\GF}{GF}$It seems that a slightly more general result holds true. This states that

for a prime $p$, if $n > 2$, or $n = 2$ and $p > 2$, then the polynomial $x^{p^{n}}-x-a$ is reducible over $\GF(p)$.

Suppose $f = x^{p^{n}}-x-a$ is irreducible over $F = \GF(p)$. (In particular, of course, $a \ne 0$.)

Then, if $\beta$ is a root of $f$, we have that $E = F[\beta]$ is a field with $p^{p^{n}}$ elements.

Since $f(\beta) = 0$, we have $$ \beta^{p^{n}} = \beta + a, \quad (\beta^{p^{n}})^{p^{n}} = \beta^{p^{2 n}} = \beta^{p^{n}} + a = \beta + 2 a, \quad \dots \quad \beta^{p^{p n}} = \beta + p a = \beta. $$ It follows that $\beta$ lies in the field $L$ with $p^{p n}$ elements, as $\beta$ is a root of the polynomial $x^{p^{p n}} - x \in F[x]$. But then $E \subseteq L$, so that $p^{n} \mid p n$, and $p^{n-1} \mid n$. This is only possible (if I'm not mistaken) for $n = 1$, and any $p$, or for $p = n = 2$.

And in fact, for $a \ne 0$, the polynomial $x^{p} - x - a$ is irreducible over $F$. And the polynomial $x^{4} - x - 1$ is irreducible over $\GF(2)$.