Solve $x^5 - x = 0$ mod $4$ and mod $5$ I'm trying to solve $$x^5-x=0$$ in $\mathbb{Z/5Z}$ and $\mathbb{Z/4Z}$
I don't see how to proceed, could you tell me how ?
Thank you
 A: Over any commutative ring:
$$x^5-x=x(x^4-1)=x(x^2-1)(x^2+1)=x(x-1)\overbrace{(x+1)}^{=(x-4)\pmod5}(x^2+1)$$
Now, as $\;5=1\pmod 4\;$ , the element $\;-1\;$ has a square root there (which in this particular case isn't surprising as $\;-1=4\pmod 5\;$) , and we have
$$x^2+1=(x-2)(x-3)\pmod 5\implies x^5-x=x(x-1)(x-2)(x-3)(x-4)\pmod 5$$
You now try the other case, but be careful: now you don't have a field.
A: In $\mathbb{Z}/5\mathbb{Z}$, the answer follows quickly from Fermat's Little Theorem.  
In $\mathbb{Z}/4\mathbb{Z}$, by the Euler Totient Theorem, we have
$$x^{\varphi(4)} \equiv x^2 \equiv 0 \mod 4$$
for $x$ coprime to $4$.  The only remaining cases are $x=0$ and $x=2$, which can easily be checked.

I observe that the above examples can be seen in broader context:  again by Fermat's Little Theorem, every element $x$ of $\mathbb{Z}/p\mathbb{Z}$ (p prime) satisfies $x^p - x = 0$ (and more generally, any element of the finite field $\mathbb{F}_{p^k}$ for all $k$).  Similarly, in $\mathbb{Z}/p^k\mathbb{Z}$, every element $x$ coprime to $p^k$ satisfies $x^{p^k+1} - x = 0$, since $\varphi(p^k) = p^{k-1}(p-1)$, and no other nonzero $x$ does, since we have for any noncoprime $x$ that $x = p^n$ for some $n\ge1$, so since $p^k+1 > k$, $p^k|x^{(p^k+1)}$.
