## Task $7 / 89$

All integers $x$ are to be determined for which $f(x)=x^{5}-x+5$ is a prime number!
It is $x^{5}-x=x\left(x^{4}-1\right)=x\left(x^{2}-1\right)\left(x^{2}+1\right) \equiv 0(\bmod 5)$. If indeed $x \neq 0(\bmod 5)$, then it is certainly either $x^{2} \equiv 1(\bmod 5)$ or $x^{2} \equiv-1(\bmod 5)$. Thus, $f(x)=x^{5}-x+5$ is a prime number exactly when $x^{5}-x=0$. The real solutions of this equation are $x_{1}=-1, x_{2}=0$ and $x_{3}=1$; for each of these values and only for these, $f(x)=5$. Other integer $x$-values yield $f(x)=5 g$ with an integer $g \neq 0$.