$$((w+x)(y+z)-(w-x)(y-z))^2=4(wy+xz)^2-4(w^2-x^2)(y^2-z^2)$$ is true and so if $(a-b)^2=4(e+f)^2-4ab$ then under what additional minimal conditions can we say $e=wy$ and $f=xz$ while $a=(w+x)(y+z)$ and $b=(w-x)(y-z)$?
At least may be $a=(\ell_1w+\ell_2x)(\ell_3y+\ell_4z)$ and $b=(\ell_1'w+\ell_2'x)(\ell_3'y+\ell_4'z)$ holds at some special $\ell_1,\ell_2,\ell_3,\ell_4,\ell_1',\ell_2',\ell_3',\ell_4'\in\mathbb Z$? Does it seem possible?
From $$((w+x)(y+z)-(w-x)(y-z))^2=4(wy+xz)^2-4(w^2-x^2)(y^2-z^2)$$ with $w'=(w+x)$, $z'=(y+z)$, $x'=(w-x)$ and $y'=(y-z)$ we have $$(w'z'-x'y')^2=(w'z'+x'y')^2-4\cdot w'y'\cdot x'z'$$ where $$(w'y'+x'z')=((w+x)(y-z)+(w-x)(y+z))=(2wy+2xz)$$ holds. We seek if $(a-b)^2=4(e+f)^2-4ab$ holds then is there a connection to $w',x',y',z'$ we can infer under some additional conditions?