Elementary number theory, sums of two squares

$$(a^2+b^2)(c^2+d^2)=(ac-bd)^2 +(ad+bc)^2.$$

Remark: let $u=a+ib$ and $v=c+id$ be complex numbers, with $a,b,c,d$ real. Then $uv=(a+ib)(c+id)=(ac-bd)+i(ad+bc)$. The identity of the answer just says that the norm of the product $uv$ is equal to the product of the norms. Everything is connected to everything else!

To elaborate a bit on the answer posted by André Nicolas:

https://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity

\begin{align} \left(a^2 + b^2\right)\left(c^2 + d^2\right) & {}= \left(ac-bd\right)^2 + \left(ad+bc\right)^2 \tag1 \\ & {}= \left(ac+bd\right)^2 + \left(ad-bc\right)^2. \tag2 \end{align}

Thus, there is more than one way to write a product of two sums of two squares as a sum of two squares.

For example, $$(1^2 + 4^2)(2^2 + 7^2) = 26^2 + 15^2 = 30^2 + 1^2.$$

(On April 6th, 2006, wrote a Wikipedia page titled "Fibonacci's identity". Here is what it looked like on April 20th: https://en.wikipedia.org/w/index.php?title=Fibonacci%27s_identity&oldid=50311836) Then that got merged into an older article, linked to above, and I made "Fibonacci's identity into a disambiguation page.)