Let $,,$ and $$ be natural numbers, $>1$ and suppose that $,$ have no common factor.
Prove that:
If $(^+^)|(^+^)$ then $|$
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1Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Feb 01 '20 at 15:46
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1Does this answer your question? $a,b,m,n$ be positive integers such that g.c.d.$(a,b)=1$ and $a^m+b^m|a^n+b^n$ , then $m|n$? – Feb 01 '20 at 15:47
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Clearly, $m\geq n$ thus let $m=nq+r$ for $0\leq r \leq n-1$ thus $$a^{m}+b^{m}\equiv a^{nq}(a^{r}+(-1)^{q}b^{r})\equiv 0\ (\mod a^{n}+b^{n})$$ thus if $q$ even $$a^{r}+b^{r}\equiv 0\ (\mod a^{n}+b^{n})$$ but which is not possible for any $0\leq r\leq n-1$ as $a>1$ and $\gcd(a,b)=1$ If $q$ is odd and $$a^{r}-b^{r}\equiv 0\ (\mod a^{n}+b^{n})$$ as $0 \leq r \leq n-1$ so , $r=0$ forced as $a\geq 2$ and $\gcd (a,b)=1$ thus $n|m$ and we are done.
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Please explain how did you put $a^m+b^m$ congruent to$ a^nq(a^r + b^r.(-1)^k ) $. I mean where did $ (-1)^k $ came from – Meth Feb 02 '20 at 07:04
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$b^{m}=b^{nq+r}=b^{nq}b^{r}=(b^{n})^{q}b^{r}=(a^{n}+b^{n}-a^{n})^{q}b^{r}=(-a^{n})^{q}b^{r}=(-1)^{q}a^{nq}b^{r}\equiv (\mod a^{n}+b^{n})$ – Feb 02 '20 at 10:43
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I would’ve smashed that like button but I just joined yesterday. Dont ve enough ‘reputation’ to do that. – Meth Feb 02 '20 at 11:27
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