Carmicheal numbers are square free

Question

Carmichael number square free

I was reading this question.can some one explain how to arrive from here In particular, $a^{n−1} \equiv 1\pmod{p^t}$, to here $a^n \equiv a\pmod{p^2}$.

Thank you

Answer 1

Multiply by $a$, then note that if $p^t | n$ where $t \geq 2$, then certainly $p^2 | n$.

So $a^{n-1} \equiv 1 \pmod {p^t} \implies a^n \equiv a \pmod {p^t} \implies a^n \equiv a \pmod {p^2}$ since $t \geq 2$.