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Problem. Let $a \geq 2$ and $a \in \mathbb{Z}$. Let $p,q$ be distinct primes. Prove that if $p^2|\frac{a^q-1}{a-1}$, then $a^{p-1} \equiv 1 \pmod {p^2}$.

Attempt. It feels like $p$ gets some property from the condition so that it has the required result. However, I could not figure out what that property is. I played around with Fermat's Little Theorem but got nowhere.

Comment. I'm an undergraduate beginner to number theory, who has only been taught basic things like Euclidean algorithm, Bezout's lemma, Fermat's Little Theorem etc. So, please do not give a hint referring to an advanced theorem. Thank you! Again, hint only please.

Bill Dubuque
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IncredibleSimon
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    You'd need to share much more about what you know. EG Do you know Lifting the Exponent lemma? – Calvin Lin Jan 24 '25 at 14:56
  • I'm an undergraduate beginner to number theory. No. I do not know what that lemma is, and we have not been taught this but some basic things like Euclidean algorithm, Fermat's Little Theorem, Bezout's lemma etc. @CalvinLin – IncredibleSimon Jan 24 '25 at 15:01
  • Sorry. I just changed it. @ThomasAndrews – IncredibleSimon Jan 24 '25 at 15:03
  • Don't put inforration in comments, edit your question. – Thomas Andrews Jan 24 '25 at 15:04
  • Just edited. @ThomasAndrews – IncredibleSimon Jan 24 '25 at 15:06
  • wilson's lemma (and maybe totient function) might help. – ness Jan 24 '25 at 15:09
  • Hint: Show that if $ a \equiv 1 \pmod{p}$, then $ p = q$ (hence reject this case). Now consider what happens when $ a \not \equiv 1 \pmod{p}, a^q \equiv 1 \pmod{p^2}$. Wishful-thinkingly, what other condition would be nice/immediate to help us conclude that $ a^p \equiv 1 \pmod{p^2}$? Then hopefully you made the right guess and go prove it. – Calvin Lin Jan 24 '25 at 15:14
  • Also, to be clear, this is almost immediate from Lifting the Exponent lemma (LTE). So I'd encourage you to provide more context about how you came across this question, esp in terms of what techniques you're supposed to be using to solve it. $\quad$ Theoretically we could break it down into the simpler theorems, but it does require a bunch of work that is essentially proving LTE, at least for this sepcific use case. – Calvin Lin Jan 24 '25 at 15:16
  • Ah, I found the "context" $\quad$ Hint: Apply the statement in your recently closed post to prove my original hint. (Notice that one of the solutions in the linked post reference LTE, which is my go-to technique upon seeing $x^n - y^n, x-y$.) – Calvin Lin Jan 24 '25 at 15:22
  • By the Lemma in the dupe: $,(\color{#c00}{q,p})!=!1:$ so $:a^{\color{#c00}q}\equiv \overbrace{1\equiv a^{\color{#c00}p(p−1)}}^{\phi(p^2),=,p(p-1)}\Rightarrow, a^{p−1}\equiv 1\pmod{!p^2}., $ Please unaccept the answer so this duplicate can be efficiently processed. $\ \ $ – Bill Dubuque Jan 24 '25 at 18:29

1 Answers1

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Hint: What does $a^q \equiv 1 \pmod{p^2}$ tell you about $q$ and $p^2$?

Apply Euler's theorem, not just Fermat's Little Theorem.

Hence, conclude that $ a^{p-1} \equiv a^{qk} \equiv 1 \pmod{p^2}$.

Original approach which is much more long-winded

Case 1: $ a \equiv 1 \pmod{p}$
Hint: Use the fact that $ \gcd ( \frac{a^n - b^n } { a-b} , a-b) = \gcd ( n \gcd(a, b) ^{n-1}, a - b ) $ to conclude then $p = q$. Reject this case as we have distinct primes.

Case 2: $ a \not \equiv 1 \pmod{p}$,
Show that $a^q \equiv 1 \pmod{p}$ and $a^q \equiv 1 \pmod{p^2}$.
Hint: How can we relate $q$ and $p$? What does $ a^ q \equiv 1$ tell you about the order of $a$?
Hence, conclude that $ a^{p-1} \equiv a^q \equiv 1 \pmod{p^2}$.

Notes

  • We didn't need that $q$ is prime.
Calvin Lin
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