Divisibility involving the product (ad+bc)(ac+bd) Prove that if $abcd|(ad+bc)(ac+bd)$, then $\frac{ac}{bd}$ is a perfect square of a rational number. $a,b,c,d$ are positive integers.
I am extremely lost on how to do this problem; any help?

Denote $(ad+bc)(ac+bd)=kabcd.$
Without loss of generality let $(a,b)=1,(c,d)=1,a/b=s,c/d=t,$
then $(s+t)(1+\dfrac{1}{st})=k.$
$$s=\frac{\pm\sqrt{\left(-k t+t^2+1\right)^2-4 t^2}+k t-t^2-1}{2 t}.$$
So $$\left(-k t+t^2+1\right)^2-4 t^2=D^2,\\
(\frac{-k t+t^2+1}{2t})^2-(\frac{D}{2t})^2=1.$$
There exist $m>n\in N,(m,n)=1$, such that 
$$\frac{-k t+t^2+1}{2t}=\pm \frac{m^2+n^2}{2mn},\\
\frac{D}{2t}=\frac{m^2-n^2}{2mn}.\\
\frac{-k t+t^2+1}{2t}=\frac{c^2-kcd+d^2}{2cd}=\pm\frac{m^2+n^2}{2mn}$$
Now $(c^2-kcd+d^2,cd)=1,(m^2+n^2,mn)=1,$ hence 
$$c^2-kcd+d^2=\pm(m^2+n^2),cd=mn.\\
-k t+t^2+1=\pm (m^2+n^2)/d^2,\\
t=mn/d^2, D=(m^2-n^2)/d^2.$$
$$\frac{ac}{bd}=st=\frac{1}{2}(\pm D-(-k t+t^2+1))=\frac{1}{2d^2}(\pm (m^2-n^2)-\pm (m^2+n^2))\\
=(m/d)^2  \quad or \quad (n/d)^2.\quad (st>0)$$
