Carmicheal numbers are square free

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

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 $$t \geq 2$$.
• @viru Just write down what it means: $a^n \equiv a \pmod {p^t} \iff a^n - a \equiv 0 \pmod {p^t} \iff p^t | \left(a^n - a\right) \iff a^n - a = p^tb$ (some $b \in \mathbb{Z}$ $\iff a^n - a = p^2 \cdot \left(p^{t-2} b\right) \implies p^2 | \left(a^n - a\right) \iff a^n - a \equiv 0 \pmod {p^2} \iff a^n \equiv a \pmod {p^2}$ – Sam Streeter Nov 3 '18 at 11:21
• If $\alpha$ is divisble by $\beta^n$, then it is also divisible by any $\beta^m$ with $m \geq n$. You can view this as a consequence of the transitive property of divisibility. – Sam Streeter Nov 3 '18 at 11:23