Maximum value of $\gcd(u^2-v^2, 2uv, u^2+v^2)$ 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?

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. 
