We have to show that : Given pairs of integers $\{a,b\}$ and $\{c,d\}$, there exists a pair of integer $\{u,v\}$ such that $(a^2+b^2)(c^2+d^2)=u^2+v^2$. Further, if $a,b,c,d$ are non-zero, and both of the sets $\{a^2,b^2\}$ and $\{c^2,d^2\}$ consist of distinct positive integers, then there are two different sets $\{u^2,v^2\}$ and $\{s^2,t^2\}$ such that $(a^2+b^2)(c^2+d^2)=u^2+v^2=s^2+t^2$.
The problem is to find geometric interpretation, and geometric proof of the above statement.
The algebraic proof is easy. I'm showing my solution using complex numbers, which I believe to have a connection with geometric interpretation. We set $z_1=a+bi,$ $\,z_2=c+di$ $($or, $z_3=b+ai,\,z_4=d+ci)$, and noting that $(a^2+b^2)(c^2+d^2)$ is simply $|z_1|^2|z_2|^2=|z_1z_2|^2$, we can choose $u=Re\{z_1z_2\}$ and $v=Im\{z_1z_2\}$. An alternative choice would be $u=Re\{z_3z_4\}$ and $v=Im\{z_3z_4\}$. For the $2$nd part, if the given conditions hold, the two choices of $(u,v)$ will be different and hence the proof.
How can find a geometric interpretation, and a geometric proof out of it? Thanks in advance.
Without your algebraic proof, I wouldn't know the geometric interpretation. So, you look for points in $\mathbb{C}$ whose real and imaginary part is an integer. Consider an integer grid.(so, you draw lines perpendicular to axes at integers)(lines described with $(a,0), a \in Z$ or $(0,b), b\in Z$. Now, that you have your grid, your starting points are some points in that grid(a point in a grid is intersection of two lines in a grid, by construction it is obvious that the complex number described by that point has real and complex value integers). Now, their product has module greater than $\sqrt{2}$ (unless some trivial special cases), which means that the circle with center 1 and radius product of two modules intersect the grid. We just need to prove that grid is intersected in at least one point. But, square described with lines $(ac,0), (bd,0), (-ac,0),(-bd,0)$ where $(a,b),(c,d)$ are initial points, has our circle as circle around it, so it is a proof:hence four points on the grid and on the circle and any of them four is your required solution.