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I need to show if $(a,b)=1,n$ is an odd positive integer then $\displaystyle \left(a+b,{a^n+b^n\over a+b}\right)\mid n.$

let $\displaystyle \left(a+b,{a^n+b^n\over a+b}\right)=d$

$\displaystyle d\mid {a^n+b^n\over a+b}=(a+b)^2(a^{n-3}-a^{n-4}b\dots+b^{n-3})-2ab(a^{n-3}\dots+b^{n-3})-ab^{n-2}$

$d\mid (a+b)$ so from the rest can I conclude $d=1$ as $(a,b)=1$?

Amzoti
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Myshkin
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    Why do you want to show it is $1$ instead of showing that it divides $n$? (which is what you said the question actually is.) – Andrés E. Caicedo Jul 26 '13 at 05:59
  • as $d=1$ so $1\mid n$ – Myshkin Jul 26 '13 at 06:02
  • Yes, if it happens that the gcd is $1$, it will certainly divide $n$. But the gcd may not be $1$, and still divide $n$, right? Why do you think the question asks what it asks? If the gcd is $1$, why not ask that directly? – Andrés E. Caicedo Jul 26 '13 at 06:05
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    Here is an example, showing that what you suggest is just false: $4$ and $7$ are relatively prime, $4+7=11$, and $11$ is odd. However, $4^{11}+7^{11}=121\times 16376207$, so the gcd of $11$ and $(4^{11}+7^{11})/11$ is not $1$. – Andrés E. Caicedo Jul 26 '13 at 06:10
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    Related : http://math.stackexchange.com/questions/247146/how-to-show-that-gcd-left-an-bn-over-a-b-a-b-right-gcdn-dn-1-a-b – lab bhattacharjee Jul 26 '13 at 06:34

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Modulo $a+b$, you have $a^kb^{n-k}\equiv (-1)^kb^n$, so $$ \frac{a^n+b^n}{a+b}=a^{n-1}-a^{n-2}b+a^{n-3}b^2\mp\cdots -ab^{n.2}+b^{n-1}\equiv n$$ so $n=\frac{a^n+b^n}{a+b}+(\ldots)\cdot (a+b)$.

Hagen von Eitzen
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