<s>[INST] Product of sums of square is a sum of squares. Given $a,b,c,d \in \mathbb{Z}$, there is $x,y$ such that
$$(a^2 + b^2)(c^2 + d^2) = x^2 + y^2$$
One can show this by considering the complex number $a + bi$ and $c+ di$, using complex properties to deduce that $x = ac - bd, y = ad + bc$ is a solution.   However, given that either $a^2, b^2$ are distinct or $c^2,d^2$ are distinct, then how can one can find nonzero solutions $x,y$?
Sorry, I am assuming $a,b,c,d$ are nonzero.
 [/INST] If $a=b=0$ or $c=d=0$, obviously we cannot do it. 
If one of (say) $a$ or $b$ is $0$, and one of $c$ or $d$ is $0$, there will be situations in which we cannot do it, for instance when $a=c=3$, $b=d=0$.
If one of $a$ or $b$ is $0$, but the other is not, and neither $c$ nor $d=0$, there is no difficulty.
Now examine the case all of $a$, $b$, $c$, $d$ are non-zero. Without loss of generality we may assume they are all positive.
Use the formulas
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2=(ad+bc)^2+(ac-bd)^2.$$
We need only check that at least one of $ad-bc$ and $ac-bd$ is non-zero. 
Suppose to the contrary that $ad=bc$ and $ac=bd$. Then (i) $a^2cd=b^2cd$ and (ii) $c^2ab=d^2ab$. Since none of $a$, $b$, $c$, $d$ is $0$, by cancellation we get $a^2=b^2$ and $c^2=d^2$.   
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