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I’m new to number theory and I’m solving questions in the textbook one by one. Here is one : If $m\geq 1$ and $n\geq2$ , which both of them are natural numbers , prove this statement:

$$(n-1)^2 | (n^m-1) \iff (n-1)| m.$$

This is my approach : I started from the left part of the statement; I used $$(a^n-b^n)=(a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1}),$$ to expand $n^m-1$ in the statement , Then used geometric series sum and I came to this : $$1|n^m-1$$

And I don’t know how to continue this .

I’m looking for a hint , and not an answer. Give me an answer and you’ll be cursed :)

Thank you in advance.

Bill Dubuque
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Negar Rezaei Nejad
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  • Continuing your approach: if $(n-1)^2\mid (n-1)(n^{m-1}+n^{m-2}+\ldots+n+1)$, then $n-1\mid n^{m-1}+n^{m-2}+\ldots+n+1$ (why?). – richrow Oct 08 '20 at 13:26

3 Answers3

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Hint 1: Use your factorization of $a^n-b^n$ to show that $$(n-1)^2\mid(n^m-1)\qquad\Leftrightarrow\qquad (n-1)\mid(n^{m-1}+n^{m-2}+\ldots+n+1).$$

Hint 2:

Divide $n^{m-1}+n^{m-2}+\ldots+n+1$ by $n-1$, as polynomials in $n$. What is the remainder?

Servaes
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You want to show that $$(n-1)\mid(n^{m-1}+n^{m-2}+\ldots+n+1)\qquad\iff\qquad (n-1)|m$$

But $n-1\equiv 0 \pmod {n-1}$, so $n\equiv 1 \pmod {n-1}.$ Now use this to evaluate$(n^{m-1}+n^{m-2}+\ldots+n+1) \pmod {n-1}.$

Alessio K
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Answer in more details:

Conder $n^{m-1}+n^{m-2} + . . . +n+1$ which has m terms and must also be divisible by $n-1$. Now add $m-1$ times (-1) and (m-1) times (+1) you get:

$(n^{m-1}-1)+(n^{m-2}-1) + . . . +n-1 +m-1+1=(n^{m-1}-1)+(n^{m-2}-1) + . . . +(n-1) +m$

All terms have factor $n-1$, so m must also have factor $n-1$ that is $n-1|m$

sirous
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