Question

If $p$ is a prime, and $x$ prime to $p$, show that $x^{p^r}-p^{r-1}-1$ is divisible by $p^r$.

Answer 1

$⟹$$x^{p-1}=M\left(p\right)$ i.e. Multiple of p

$x^{p-1}=1+k_p$ ( k is any constant)

$\therefore {\left(x^{p-1}\right)}^{p^{r-1}}={(1+k_p)}^{p^{r-1}}=1+k_{p}M\left(p^{r-1}\right)$

$x^{p^r-p^{r-1}}=1+k_{p}M\left(p^{r-1}\right)$

$x^{p^r-p^{r-1}}-1=M\left(p^r\right).$