Divisibility involving the product (ad+bc)(ac+bd)

Question

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?

Answer 1

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 $$c^2-kcd+d^2=\pm(m^2+n^2),cd=mn.\\ -k t+t^2+1=\pm (m^2+n^2)/d^2,\\ t=mn/d^2, D=(m^2-n^2)/d^2.$$ $$\frac{ac}{bd}=st=\frac{1}{2}(\pm D-(-k t+t^2+1))=\frac{1}{2d^2}(\pm (m^2-n^2)-\pm (m^2+n^2))\\ =(m/d)^2 \quad or \quad (n/d)^2.\quad (st>0)$$