Show that $M_p^p\equiv 1 \mod p^2$

Question

Can it be shown that $M_p^p\equiv 1 \mod p^2$ where $M_p=2^p-1$ is a Mersenne prime.

I tried to develop the left part into into $2^{p^2}-1-pk2^p$ and use $2^{p^2}\equiv 2^p \mod p^2$, but I get nowhere

Thanks

Answer 1

From FLT $$2^{p-1}\equiv 1\pmod{p} \Rightarrow 2^{p}\equiv 2\pmod{p} \Rightarrow 2^p-1\equiv 1 \pmod{p} \Rightarrow$$ $$M_p \equiv 1 \pmod{p} \tag{1}$$ And from $$M_p^p-1=\color{red}{(M_p-1)}\color{blue}{(M_p^{p-1}+M_p^{p-2}+M_p^{p-3}+...+M_p+1)}$$ we have

• the expression in red is divisible by $p$, from $(1)$
• also from $(1)$ follows $$M_p^{p-1} \equiv 1\pmod{p}$$ $$M_p^{p-2} \equiv 1\pmod{p}$$ $$..$$ $$M_p \equiv 1\pmod{p}$$ $$1 \equiv 1\pmod{p}$$ if we sum them $$M_p^{p-1}+M_p^{p-2}+M_p^{p-3}+...+M_p+1\equiv \underbrace{1+1+1+...+1}_{p\text{ times}}=p\equiv 0\pmod{p}$$thus, the expression in blue is also divisible by $p$.

As a result $p^2 \mid M_p^p-1$.