Elementary number theory, sums of two squares Suppose x can be written as a sum of two squares of integers, and y can be written as a sum of two squares of integers. Show that xy can also be written as a sum of two squares of integers.
 A: 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.)
A: $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2 +(ad+bc)^2.$$
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! 
