Note that \[(a^2+b^2)(c^2+d^2)=(ad+bc)^2+(ac-bd)^2=(ab+cd)^2\]and also \[(a^2+d^2)(b^2+c^2)=(ab+cd)^2+(ac-bd)^2=(ad+bc)^2.\]Adding them together, we'll get $ac=bd$ and $ad+bc=ab+cd \iff a(d-b)=c(d-b) \iff (a-c)(d-b)=0$. Now, wlog, $a=c$, so we get $a^2=bd$. Now, since $gcd(a,b,c,d)=1$, we must have $gcd(b,d)=1$ (it's easy to be proven by contradiction), which implies $b=k^2$, $d=m^2$, and $a=c=km$ for some $k,m \in \mathbb{N}$. Finally, we have $a+b+c+d=km+k^2+km+m^2=(k+m)^2$, which is solved.