The explicit monoid law on the differences of two square integers?

Question

It is known that if $S = \{ x^2 : x \in \Bbb{Z}\}$ is the submonoid, then $\Delta S = \{ x^2 - y^2 : x,y\in \Bbb{Z}\}$ or the set of all differences of squares forms the monoid $4\Bbb{Z} \uplus( 2\Bbb{Z} + 1)$.

I was wondering, can you take two $a^2 - b^2, c^2 - d^2$ and come up with a formula for $x,y$ in terms of $a,b,c,d$ such that $(a^2 - b^2)(c^2 - d^2) = x^2 - y^2$?

Answer 1

$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$ is a famous identity. Now substitute $b=ib’,d=id’$ to obtain $$(a^2-b^2)(c^2-d^2)=(ac-bd)^2-(ad-bc)^2$$.