Question

If $p$ is a prime number, $x^p-x$ is divisible by $p$.

Answer 1

Let $x^p-x$ be denoted by $f\left(x\right)$; then
$f(x+1)-f\left(x\right)=(x+1{)}^p-(x+1)-(x^p-x)$
$=p{x}^{p-1}+\frac{p(p-1)}{1.2}{x}^{p-2}+....+px$
$=$ a multiple of $p$, where $p$ is prime
$\therefore f(x+1)=f\left(x\right)+$ a multiple of $p$.
If $f\left(x\right)$ is divisible by $p$, so also is $f(x+1)$; but
$f\left(2\right)=2^p-2=(1+1{)}^p-2$,
and this is a multiple of $p$ where $p$ is prime; therefore $f\left(3\right)$ is divisible by $p$, therefore $f\left(4\right)$ is divisible by $p$, and so on; thus the proposition is true universally.