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This question already has an answer here:

Finding my previous question quite naive, I improve my question:

Given that $n,a,b \in \mathbb{N}$ and $n\mid(a^n-b^n)$ , can we prove or disprove $n\mid(a^n-b^n)/(a-b)$ ?

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marked as duplicate by Zander, Claude Leibovici, M. Vinay, user91500, loup blanc Jun 29 '16 at 8:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @Roby5 Would you like to elaborate on this comment, as it seems that the claim is valid. $\endgroup$ – Stefan4024 Jun 28 '16 at 15:16
  • $\begingroup$ Can we really set the value of n in this way? $\endgroup$ – Linear Jun 28 '16 at 15:18
  • $\begingroup$ @Linear To be fair I don't see how Roby5's suggestion helps us. I mean after all it doesn't disprove the claim and it's doing nothing to prove it. $\endgroup$ – Stefan4024 Jun 28 '16 at 15:20
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Let $p$ be an odd prime divisor of $n$. And let $v_p(n)$ denote the highest power of $p$ in the prime decomposition of $n$. Now assume that $v_p(n) = k$

Now if $p \not \mid a-b$ then as $p^k \mid (a^n - b^n)$ we have that $p^k \mid \frac{a^n - b^n}{a-b}$

If $p \mid a-b$ and $p \not \mid a$ then we have that $p \not \mid b$. Now using the following lemma we have:

$$v_p\left(\frac{a^n - b^n}{a-b}\right) = v_p(a^n - b^n) - v_p(a-b) = v_p(a-b) + v_p(n) - v_p(a-b) = v_p(n) = k $$ $$\implies p^k \mid \frac{a^n - b^n}{a-b}$$

If $p \mid a-b$ and $p\mid a$ then we have that $p \mid b$. Let $m$ be the highest power of $p$ dividing them both. Then we have:

$$\frac{a^n - b^n}{a-b} = p^{m(n-1)}\frac{(a_1)^n - (b_1)^n}{a_1-b_1}$$

Now obviously $p^k \mid \frac{a^n - b^n}{a-b}$, as $k \le m(n-1)$

NOTE: If $p=2$ only the second part has to be altered, but that's easy again using the lemma mentioned above.

Now since for every prime divisor of $n$ we have that $p^k \mid \frac{a^n - b^n}{a-b}$ we have that $n \mid \frac{a^n - b^n}{a-b}$. Q.E.D.

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