Roots of $x^p + x + [\alpha]_p \in \mathbb{F}_p[x]$

Question

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?

Answer 1

As you note, by Euler we have $x^p\equiv x\pmod p$ for all $x\in\Bbb{Z}_p$, so $$x^p+x+[\alpha]_p\equiv 2x+[\alpha]_p,$$ holds for all $x\in\Bbb{Z}_p$. So if $p\neq2$ then $x=-\tfrac{1}{2}[\alpha]_p$ is a root, and it is in fact the only root.

Answer 2

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:

1. 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 $\alpha$.
2. If $p=2$, then you already have the solution.