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How to show $(x^m+1,x^n+1)=x^{(m,n)}+1$, where $m,n$ are odd numbers.

It is easy to see that $x+1$ are divisors of $x^k+1, k=m,n$. But how to prove it is the greatest one? It sounds as follows. there exists integers $u,v$ such that $um+vn=d$, where $d=(m,n)$. Assume w.l.o.g, $u<0, v>0$, then $x^{-um}(x+1)=x^{vn}+x^{-um}=x^{vn}+1-1+x^{-um}$. Oh. $x^{vn}+1$ can be divided by $x^n+1$, But how to treat $x^{-mu}-1$?

Here, $(a,b)$ is the greatest common divisor.

xldd
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  • Does (a,b) mean greatest one among $a$ and $b$? – Ilovemath Sep 06 '21 at 09:55
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    @Ilovemath greatest common divisor – xldd Sep 06 '21 at 09:56
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    Check this: https://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1. – Martin R Sep 06 '21 at 09:58
  • Please, make up your mind. In the title you wrote $(x,y)=1$, then that hypothesis is not in the body. Then you call $d$ the gcd of $m$ and $n$, but then you don't use $d$, you used $um+vn=1$. Are or are not $m$ and $n$ coprime? – jjagmath Sep 06 '21 at 10:16
  • This is wrong: $x^{vn}+x^{-um}=x^{vn}+1+1-x^{-um}$. – jjagmath Sep 06 '21 at 10:19
  • It seems to me that these answers https://math.stackexchange.com/a/1287103/42969, https://math.stackexchange.com/a/1464115/42969 (with $b=-1$) applies directly to your problem. I would therefore suggest to close the question as a duplicate. – Martin R Sep 06 '21 at 11:07

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