# diophantine equations – Are there integers nonzero integers \$a,b,c,d,x,y,m,n,p\$ such that \$(a^2-mb^2)(c^2-nd^2)=x^2-py^2\$? (\$m,n,p)\$ are square free non equal integers.

We are all familiar with Fibonacci-Brahmagupta’s identity:

$$(a^2-mb^2)(c^2-md^2)=(ac+ mbd)^2-m(ad+bc)^2$$
I am trying to find whether there is a similar identity:
$$(a^2-mb^2)(c^2-nd^2)=x^2-py^2$$ where $$p,m,n$$ are not all equal.

If this problem has already been solved, please save me the trouble by pointing me at a reference. Otherwise, any hint will be greatly appreciated.Thanks.