<s>[INST] Proof-verification: of numbers such that $(a^2+b^2)(c^2+d^2)=u^2+w^2$ I would like to ask if someone can verify that my solution to the following question is correct:
Question:
Prove that for any $a,b,c,d\in\mathbb{Z}$ there exists integers $u,v$ such that
$$(a^2+b^2)(c^2+d^2)=u^2+v^2$$
My solution:
Let $z_1 = a+ib$ and $z_2 = c+id$ then one has that
$$(a^2+b^2)(c^2+d^2)=|z_1|^2|z_2|^2=|z_1z_2|^2\\
=|(ac-bd)+(ad+bc)i|^2\\
=(ac-bd)^2+(ad+bc)^2$$
Let $u=ac-bd$ and $v=ad+bc$. And since $a,b,c,d\in\mathbb{Z}$, $u,v\in\mathbb{Z}$. QED
My Question: Is this correct? And if not, where is my mistake?
 [/INST] There are an infinite number of solutions where the right side is a hypotenuse and the left side is the product of hypotenuses of smaller Pythagorean triples. One example is
$$(3,4,5)\rightarrow 3^2+4^2=5^2\quad (5,12,13)\rightarrow 5^2+12^2=13^2\\
5\times13=65\quad \implies\quad  a=2, b=1, c=3, d=2\\
\implies u=7, v=4\quad\lor\quad u=8, v=1$$
There will be $2^{n-1} $ solutions where $n$ is the number of unique prime factors of $u^2+v^2$  provided that  these [prime] factors are of the form $4n+1.$ In this example the $2^{2-1}=2$ prime factors were $5$ and $13$.
Other solutions are
$$a=2, b=1, c=4, d=1\qquad  u=7, v=6\quad\lor\quad u=9, v=2$$
$$a=2, b=1, c=5, d=2\qquad  u=9, v=8\quad\lor\quad u=12, v=1$$
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