Prove that every number in $\mathbb{Z}$ is a solution to the congruence $$x^7 − 2x \equiv x \ \ (\operatorname{mod} 42)$$
As far as I can see, this congruence does not have any solutions (for example if we take $x=3$, the output is incorrect), but the book says I need to prove otherwise. Am I missing something here?
Yes, the given congruence has solutions, and you are right, $x=3$ is not one of them.
In order to find them, since $42=2\cdot 3\cdot 7$, it follows that the congruence is equivalent to the system $$\begin{cases} x^7 − 3x \equiv 0 &\pmod{2}\\ x^7 − 3x \equiv 0 &\pmod{3}\\ x^7 − 3x \equiv 0 &\pmod{7} \end{cases} \Leftrightarrow \begin{cases} x-x \equiv 0 &\pmod{2}\\ x -0\equiv 0 &\pmod{3}\\ x-3x \equiv 0 &\pmod{7} \end{cases}$$ because, by the Fermat's little theorem, $x^p\equiv x \pmod{p}$ for any prime $p$. Can you take it from here?