If $a,b,c,d\in\mathbb N$ and $a^2+b^2\mid ac+bd$, can it be true that $\gcd(a^2+b^2,c^2+d^2)=1$? or $3$? or $74$?
That problem is complicated. I've tried some approaches, but they're useless. E.g. if $$\gcd(a^2+b^2,c^2+d^2)=1$$ Then $$\gcd((a^2+b^2)(c^2+d^2),ac+bd)=\gcd(a^2+b^2,ac+bd)\gcd(c^2+d^2,ac+bd)$$ So it would be sufficient to show that this equality can't hold. But it won't work.
Also, if $\gcd(a,b)=1$, then $\gcd(a+b,a^2+b^2)=1$ or $2$. So it'd also be sufficient to prove that it can't be true that $$\gcd(a^2+b^2+c^2+d^2,(a^2+b^2)^2+(c^2+d^2)^2)$$ is $1$ or $2$, but you can see how desperate proving this would actually be.
I'm curious to see a solution. I'll see if you could think of one.
And it's a problem from the selection of people for the IMO $2013$ phase (not sure how to say it). I can't find a solution on the Internet. I think knowing how to solve this problem could help me in the future, so I've posted it here.