Positive rational numbers with sum of their squares equal to $n$ 
Prove that for all positive integers $n$ there exist $n$ distinct, positive rational numbers with sum of their squares equal to $n$.

For $n = 1$ we can just take $1$. For $n = 2$ we can take $\left(\dfrac{1}{5}\right)^2+\left(\dfrac{7}{5}\right)^2 = 2$. How can we generalize this to any $n$?

You can get $x^2+y^2=1$ infinitely many ways using Pythagorean triples, just make sure all the triples are distinct, then add the expressions. If $n$ is odd, use an initial $1$ not from a triple.
Edit 2: (previous idea didn't work) Suppose that
$$x=\frac{k^2-2k-1}{k^2+1},\ y=\frac{k^2+2k-1}{k^2+1}. \tag{1}$$
Then $x^2+y^2=2,$ and if $k$ is even both $x,y$ in (1) can be shown to be in lowest terms. This means we have infinitely many distinct solutions to $x^2+y^2=2,$ so that as @N.S suggests in his comment below we can get $n$ distinct rational squares with sum $n.$
