$(x^m+a^m,x^n+a^n)=x^d+a^d$, for complex $a\neq 0$

Asked Sep 08 '21 at 07:41

Active Sep 08 '21 at 07:41

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$(x^m+a^m,x^n+a^n)=x^d+a^d$, for complex $a\neq 0$, where $m,n$ are postive integers, $d$ is the gcd of $m,n$, and $m/d, n/d$ are both odd numbers.

Clearly, let $m=m_1d, n=n_1d$, then $m_1,n_1$ are odd numbers, so $x^m+a^m=(x^{m_1d}+a^{m_1d}) =(x^d+a^d)(\cdots)$. So $x^d+a^d$ is a divisor of $x^m+a^m, x^n+a^n$. But how to prove it is greatest common divisor? Any hint would be appreciated.

asked Sep 08 '21 at 07:41

xldd

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Rescaling (i.e. using the automorphism $P(x)\Leftrightarrow P(\frac{x}{a})$ for polynomials $P$), you can assume $a=1$. – Ewan Delanoy Sep 08 '21 at 07:46
Perhaps$(\Phi_n,\Phi_m)=\Phi_{(m,n)}$ (where $\Phi_n$ is the $n$-th cyclotomic polynomial) can be used here. – Ewan Delanoy Sep 08 '21 at 07:48

$(x^m+a^m,x^n+a^n)=x^d+a^d$, for complex $a\neq 0$

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