Let $a,b,m,n$ be positive integers such that g.c.d.$(a,b)=1$ and $a^m+b^m\mid a^n+b^n$, then is it true that $m$ divides $n$?
$a,b,m,n$ be positive integers such that g.c.d.$(a,b)=1$ and $a^m+b^m\mid a^n+b^n$ , then $m\mid n$?
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1Have you some empirical evidence. context, intuition, etc? It would be easier if we "knew" if it is true or not. Furthermore, if it is a problem from a textbook, we can guess that the difficulty is not overwhelming. – ajotatxe Dec 28 '16 at 11:55
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Have you found numbers such that $a^m+b^m\mid a^n+b^n$ and $n>m\ge 2$? – ajotatxe Dec 28 '16 at 12:03
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Wrong paste . see http://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 – lab bhattacharjee Dec 29 '16 at 17:00
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Let $n\equiv s\pmod m$, with $0\le s<m$, and $n=mk + s$. We have that
$$a^n + b^n \equiv b^{mk}((-1)^ka^s + b^s)\equiv 0 \pmod {a^m + b^m}$$ but $(b^{mk},a^m+b^m)=1$, so $$(-1)^ka^s + b^s\equiv 0 \pmod {a^m + b^m}$$ but $$|(-1)^ka^s + b^s|\le a^s + b^s < a^m+b^m $$ so $(-1)^ka^s + b^s=0$. $a,b$ are coprime, so we have necessarily $$a^s=b^s=1\implies s=0\implies m|n$$
Edit:
There is only one exception: if $a=b=1$ then any $m,n$ are ok, so the statement is true if $ab>1$.
Exodd
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1Note for first congruence: $$b^{mk}(-1)^ka^s\equiv(a^m-a^m+b^m)^k(-1)^ka^s\equiv(-a^m)^k(-1)^ka^s\equiv a^{mk}a^s\equiv a^n$$ – ajotatxe Dec 28 '16 at 12:14
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