What is the maximum of $\gcd(u^2-v^2, 2uv, u^2+v^2)$, where $u$ and $v$ are integers and $u\neq v\neq 0$ ? I have tried to find it but I haven't got anywhere. I got this question when I was watching this video, where he says, at 9:36, that we need not scale down by less than $1/2$ (you might want to rewind the video). How do we prove it?
Follow up question: What about $\gcd(u^2-v^2-w^2, 2uv, 2uw, u^2+v^2+w^2)$ and so on?
On
One must of course suppose that $u, v$ are coprime. Obviously $d\mid u^2+v^2$ and $2uv$ iff $d\mid(u+v)^2$ and $(u-v)^2$. If moreover $d$ is a prime $p$ , the condition becomes $p \mid(u+v)$ and $(u-v)$, so that a prime common divisor of $u^2-v^2, 2uv, u^2+v^2$ is just a prime common divisor of $(u+v)$ and $(u-v)$, which must be $2$ because $u, v$ are coprime. So your gcd must be a power $2^n$. If $n>1$, then $2$ would divide e.g. $u$ in the product $uv$, but would not divide $u^2\pm v^2$ : impossible.
As for the "generalization" you ask for, the previous argument could work only if you add expressions of the form $u^2\pm v^2\pm w^2$ in order not to break symmetry.
Let $d=\gcd(u,v)$, $u=dx$, $v=dy$. Then $$\gcd(u^2-v^2,2uv,u^2+v^2)=d^2\gcd(x^2-y^2,2xy,x^2+y^2).$$ Now if $p\mid x$ then $p\nmid x^2+y^2$ and if $p\mid y$ then $p\nmid x^2+y^2$. Hence $\gcd(x^2-y^2,2xy,x^2+y^2)$ can at most be $2$, and this happens iff $x,y$ are both odd.