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 $74$? 
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.
Thanks.

We have the identity
$$
 b^2\bigl((a^2+b^2)-(c^2+d^2)\bigr) = (a^2+b^2)(c^2-b^2) - (ac-bd)(ac+bd).
$$
Hence $(a^2+b^2) \mid (ac+bd)$ forces $(a^2+b^2) \mid b^2(c^2+d^2)$. Since $a^2+b^2 > b^2$ for all $a,b \in \mathbb{N}$, we must have $\gcd(a^2+b^2,c^2+d^2)>1$. Furthermore, $\gcd(a^2+b^2,b^2)$ must divide $a^2\!$, so any prime $p$ involved must divide $\gcd(a,b)$, and hence $p^2 \mid (a^2+b^2)$.
Can you take it from there?
