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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

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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 $t \geq 2$.

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    $\begingroup$ @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 $b \in \mathbb{Z}$ $\iff a^n - a = p^2 \cdot \left(p^{t-2} b\right) \implies p^2 | \left(a^n - a\right) \iff a^n - a \equiv 0 \pmod {p^2} \iff a^n \equiv a \pmod {p^2}$ $\endgroup$ – Sam Streeter Nov 3 '18 at 11:21
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    $\begingroup$ Thank you for you patience .Should have done it myself.Got it :) $\endgroup$ – viru Nov 3 '18 at 11:23
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    $\begingroup$ 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. $\endgroup$ – Sam Streeter Nov 3 '18 at 11:23
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    $\begingroup$ @viru No problem! Good luck with your studies in number theory, hope you enjoy it as much as I do! $\endgroup$ – Sam Streeter Nov 3 '18 at 11:26
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    $\begingroup$ @viru One more thing: to format mod notation, you should use "pmod" (with the backslash). This fills in the brackets for you as well as making it look right, so you just need to put pmod and then the modulus you're working over (in curly brackets if it's more than one character, has a power etc.). $\endgroup$ – Sam Streeter Nov 3 '18 at 11:28

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