I’m super new to number theory and here is a question from a textbook:
Prove this : if $1\le m$ and $2\le n$ ,m and n are natural numbers then we have : $$(n-1)^2|(n^m-1) \iff (n-1)|m$$
I even appreciate a hint or guiding me through the answer.
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1Hey! What have you tried so far? Are you sure that the greater or equal sign is in the correct direction? – Vinyl_cape_jawa Sep 22 '20 at 06:16
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https://math.stackexchange.com/questions/3535664/an-assignment-problem-in-elementary-number-theory-course-on-which-i-am-struck – lab bhattacharjee Sep 22 '20 at 06:52
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Factorize the value of $n^m-1$ as below: $$(n-1)(n^{m-1}+n^{m-2}+\cdots+n+1)$$ Now, the second factor has to be divisible by $(n-1)$. It has $m$ terms. All of them are $1 \bmod{(n-1)}$. Can you complete the proof? Can you use this same factorization to show the converse?
Haran
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