The product of a sum of squares is a sum of squares If $x=a^2+b^2$ and $y=c^2+d^2$ 
how can i prove that xy is also the sum of two rational squares? 
My teacher told me there are various methods to attack this problem but an easy way is to use the norms of guassian integers but I can't see how it would help. 
 A: The idea here is to do as your teacher said, write explicitly the gaussian integers norm and then rearrange the product to end up with the norm of another gaussian integer.
$$
xy=(a^2+b^2)(c^2+d^2)\\
  =(a+ib)(a-ib)(c+id)(c-id)\\
  = (a+ib)(c-id)(a-ib)(c+id)\\
= (ac-iad+ibc+bd)(ac+iad-ibc+bd)\\
=((ac+bd)+i(bc-ad))((ac+bd)-i(bc-ad))\\
=(ac+bd)^2+(bc-ad)^2
$$
A: Since you mentioned the use of Gaussian integers specifically: (and assuming that $a,b,c,d$ are supposed to be integers)
We define the norm on the ring of Gaussian integers as follows:
$$N: \mathbb{Z} + i\mathbb{Z} \to \mathbb{Z}: a + ib \mapsto (a+ib)(a - ib) = a^2 + b^2.$$
Since $x = a^2 + b^2$, we have that the element $a + ib$ has norm $x$. Analogously, the element $c + id$ has norm $y$. Now we have that $(a+ib)(c + id)$ is an element of the ring of Gaussian integers (by definition of a ring), hence there exists elements $e,f \in \mathbb{Z}$ such that 
$$(a+ib)(c + id) = e + if$$
and hence their norms are equal. The norm is a multiplicative map, so we find that 
$$xy = N(a+ib)N(c + id) = N((a+ib)(c + id)) = N(e + if) = e^2 + f^2.$$
