$x=\frac{2-2t^2}{1+t^2}$, $y=\frac{4t}{1+t^2}$. Are you sure the problems you are posting are from a real TST ?

I took part in this TST

This is too easy for TST even for such country as Kosovo.

math_mat wrote: $x=\frac{2-2t^2}{1+t^2}$, $y=\frac{4t}{1+t^2}$. What is the motivation for this construction?

PiAreSquare wrote: What is the motivation for this construction? Pythagorean tripples : write $x=2\frac ac$ and $y=2\frac bc$ and equation is $a^2+b^2=c^2$

yeah all you need to do is state that there exist an infinite number of Pythagorean Triples (which is a well known fact) then the proof is trivial

Generalization:Let $k\in\mathbb{N^*}$ be a positive integer.Then the equation $x^2+y^2=k^2$ has infinitely many rational solutions. Solution:By Euclid's Formula we have $(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2,\forall m,n\in\mathbb{N^*}$,so we can take $x=\frac{k(m^2-n^2)}{m^2+n^2},y=\frac{2kmn}{m^2+n^2}$ for any two positive integers $m,n$ and we will obtain $x^2+y^2=(\frac{k(m^2-n^2)}{m^2+n^2})^2+(\frac{2kmn}{m^2+n^2})^2=k^2\cdot\frac{(m^2-n^2)^2+(2mn)^2}{(m^2+n^2)^2}=k^2$.