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How can we prove that $\gcd(m^2,m+n) = \gcd(n^2,m+n)$?

My try:
$$ \gcd(m^2,m+n) = \gcd\left[m^2=m\cdot \left(m+n\right)-mn\right] = \gcd(m+n,mn)$$
But I don't think it'll help me for the future, there is another way maybe?

Sammy Black
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Xavi
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    Why it doesn't help you for the future? It's a good solution, I think. – WhatsUp Dec 08 '21 at 23:22
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    @Xavi Note that the central term doesn't make sense. Can you fix it to display what you want? – Calvin Lin Dec 08 '21 at 23:32
  • By [gcd mod reduction](https://math.stackexchange.com/a/95825/242) in the linked dupe: $\,\gcd(m^2,m+n) = \gcd(n^2,m+n)\,$ by $\,m^2\equiv n^2\pmod{m+n},\,$ since $\,m+n\mid m^2-n^2,\,$ or, more *arithmetically* by $\!\bmod m+n\!:\ m\equiv -n\Rightarrow m^2\equiv (-n)^2\equiv n^2.\ $ More generally this shows $\gcd(f(m),m+n) = \gcd(f(-n),m+n)\,$ for any polynomial $f(x)$ with *integer* coefs, since the [Polynomial Congruence Rule](https://math.stackexchange.com/a/879262/242) $\Rightarrow f(m)\equiv f(-n)\pmod{m+n}\ \ $ – Bill Dubuque Dec 09 '21 at 03:22

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As @WhatsUp said, you can continue this to a solution.

$$ \gcd(m+n,mn)=\gcd(m+n,-mn)$$ $$=\gcd(m+n,-mn+n(m+n))=\gcd(m+n,n^2)$$

Robert Shore
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Addem
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    Or more directly from what OP has, similarly $ \gcd(n^2, m+n) = \gcd(m+n, mn) $, so these values are equal. (I'm guessing that's why @WhatsUp said "it's a good solution" as opposed to "It's a good start".) – Calvin Lin Dec 08 '21 at 23:30
  • Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement [here](https://math.meta.stackexchange.com/q/33508/242). – Bill Dubuque Dec 09 '21 at 03:27