TITLE: Roots of $x^p + x + [\alpha]_p \in \mathbb{F}_p[x]$
QUESTION [6 upvotes]: Let $$g(x) = x^p + x + [\alpha]_p \in \mathbb{F}_p[x],$$ where $p$ is prime.

For which $\alpha, p \in \mathbb{Z}$ does $g(x)$ have at least
one root? And for which $\alpha, p \in \mathbb{Z}$ does $g(x)$
have exactly  one root?

Supposing that $\beta \in \mathbb{Z}$ is a root of $g(x)$, then, by Euler's Theorem (or Fermat's Little Theorem), I get that it is a root of $$h(x) = 2x + [\alpha]_p.$$ So we can say that for $p=2$ we have two roots (that is $0,1$) if $\alpha \equiv 0 \mod 2$ and no root if $\alpha \equiv 1 \mod 2$. What can we conclude for the other cases?

REPLY [2 votes]: Consider
$$\gcd(x^p+x+\alpha,x^p-x)=\gcd(2x+\alpha,x^p-x),$$
where $\gcd$ denotes the greatest common divisor in $\mathbb Z_p$. Observe that, if the degree of the gcdis positive, then it is equal to the number of distinct roots of $g(x)$ in $\mathbb Z_p$. 
We have two cases:

If $p\neq 2$ then $$\gcd(2x+\alpha,x^p-x)=x+2^{-1}\alpha,$$ and $-2^{-1}\alpha$ is the unique root in $\mathbb Z_p$, for every $\alpha$.
If $p=2$, then you already have the solution.