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How does one prove that if $n \mid (a^{n}-b^{n}) \ \Longrightarrow$ $ \displaystyle n \mid \frac{a^{n}-b^{n}}{a-b}$ where $a,b, n \in \mathbb{N}$.

What i thought of is to consider $(a-b)^{n} \equiv a^{n} + (-1)^{n}b^{n} \ (\text{mod} \ n)$ and if we suppose that $n$ is odd then we have, $(a-b)^{n} \equiv a^{n} -b^{n} \ (\text{mod} \ n)$ and since $n \mid (a^{n} - b^{n})$ we have $(a-b)^{n} \equiv 0 \ (\text{mod} \ n) $

I think i am far away from the conclusion of the problem, but this is what i could work on regarding the problem.

  • 1
    The l$a$st line of your work is incorrect. ($a$-b)^{n} \equiv 0 (mod n) does not imply n \mid (a-b). Take a = 6, b = 4, n = 4 for a counterexample.2010-10-08

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Use the fact that $x \equiv y \pmod{p^{\ell}}$ (with \ell > 0) implies $x^p \equiv y^p \pmod{p^{\ell+1}}$ to treat the case where $n$ is a prime power. But note that to prove the statement for $n$, it suffices to prove the statement separately for each prime power dividing $n$. [I am of course leaving out details, but perhaps the OP might enjoy trying to fill in the sketch.]