$n|a^n-b^n\Rightarrow n|\frac{a^n-b^n}{a-b}$

Question

Let $n\in\mathbb{N}$ and $(a,b)\in \mathbb{Z}^2$. Show that: $$n|a^n-b^n\Longrightarrow n|\frac{a^n-b^n}{a-b}$$

I’ve tried an induction, but I gave up. Is there a direct proof?

To admin: Please open this post and wait one day before closing it since we are looking for new perspectives.

I could imagine the converse. – Wuestenfux Feb 12 '19 at 16:54 — Wuestenfux, Feb 12 '19 at 16:54
The claim is obviously true for $n=2$. – daw Feb 12 '19 at 17:05 — daw, Feb 12 '19 at 17:05
Note $n$ acts both as divisor and as exponent. – daw Feb 12 '19 at 17:07 — daw, Feb 12 '19 at 17:07

nonuser · Accepted Answer · 2019-02-12T17:12:45.177

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If $n$ is prime then it is true. Since $$a^n-b^n ={a^n-b^n\over a-b}\cdot (a-b)$$

$n$ must divide one of them. If it divide the fraction we are done. Suppose it divide $a-b$, then

$${a^n-b^n\over a-b} =a^{n-1}+a^{n-2}b+...+b^{n-1}\equiv n\cdot a^{n-1} \equiv 0\pmod n$$

and we are done again.

edited Feb 12 '19 at 17:12

answered Feb 12 '19 at 16:58

nonuser

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What about when $n$ isn't a prime? – Dr. Mathva Feb 12 '19 at 17:15
Why $n$ must divide one of them? – DINEDINE Feb 12 '19 at 17:16
By Euclid's lemma (since $n$ is a prime) – Dr. Mathva Feb 12 '19 at 17:16
12|24=3x8 but 12 dont divide 3 or 8 – DINEDINE Feb 12 '19 at 17:17
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$12$ isn't prime. – fleablood Feb 12 '19 at 17:18
I misread it. I have not seen that n was prime in the post – DINEDINE Feb 12 '19 at 17:19
... but why does $a^{n-1} + .... + b^{n-1} \equiv n a^{n-1} \pmod n$? Are we assuming $a^{n-1-k}b^{k}\equiv a^{n-1} \pmod n$? Because $a^n\equiv b^n$ so $a^{n-1-k}b^{k}\equiv a^{k+1}b^{n-k}$? But... what... – fleablood Feb 12 '19 at 17:26
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Because $a\equiv b$ – nonuser Feb 12 '19 at 17:29
@Aqua Can you please tell how to approach the case when n is not prime? – Jan 29 '21 at 10:05
I don't know, I have only a solution when it is prime. You better look at the linked answers. @No-One – nonuser Jan 29 '21 at 10:07

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