I really need help with this problem.
I think I should take first $d=\gcd(m,n)$
But I don't know how to use this fact. I would really appreciate some help
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1Do you mean $m, n \in \mathbb{N}$? – lhf May 24 '19 at 12:11
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Now that I see it doesn't say anything about m and n in my book – J.Dane May 24 '19 at 12:39
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1See here for a simple rigorous proof that works in any commutative ring. – Bill Dubuque May 24 '19 at 14:28
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I assume you mean $m, n \in \mathbb{N}$, and you're referring to the polynomial $\gcd$. We claim the following answer: if $m$ and $n$ have the same highest power of $2$, then $\gcd(x^m+a^m, x^n+a^n) = x^d+a^d$ where $d = \gcd(m,n)$, and a constant polynomial otherwise.
To prove this, we first replace $x$ with $ax$ to reduce the problem to $\gcd(x^m+1, x^n+1)$. We now need to find common roots of both polynomials. The roots of $x^m+1 = 0$ take form $e^\frac{(2k+1)i\pi }{m}$, which yields the set of arguments $\{\frac{\pi}{m}, \frac{3\pi}{m}, \ldots, \frac{(2m-1) \pi}{m} \}.$ We have a similar set for $n$.
If $m$ and $n$ have a different exponent of $2$ in their prime factorisation, since the numerator is an odd multiple of $\pi$, one set will have fractions with a higher factor of $2$ on the denominator than the bottom than the other, which cannot be simplified, and so there are no common roots. Otherwise, it is not too difficult to see that $\frac{(2kd-1) \pi}{d}$ are the only common elements in both sets, thus proving the statement.
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