Gcd problem. Prove that $\gcd(n + m, 2n) = \gcd(n + m, 2m)$.

Question

Suppose that $n$ and $m$ are integers and that at least one of them is nonzero. Prove that $\gcd(n + m, 2n) = \gcd(n + m, 2m)$.

Answer 1

Use the properties of Greatest common divisor: $$\gcd(n+m,2n)=\gcd(n+m,2n-2(n+m))=\gcd(n+m,-2m)=\gcd(n+m,2m).$$

Answer 2

Recall by Euclid that $\ \gcd(a,b') = \gcd(a,b)\ $ when $\,\ {b'\equiv b}\pmod{a}$

In particular, if $\,f(x)\,$ is a polynomial with integer coefficients then

$\quad \gcd(m\!+\!n,f(m)) = \gcd(m\!+\!n,f(-n))\ $ by $\ {\rm mod}\ m\!+\!n\!:\,\ m\equiv -n\,\Rightarrow\,f(m)\equiv f(-n)\,$

by the Polynomial Congruence rule. Yours is the special case $\,f(x) = 2x.$