Can someone clarify Example I.I.2 from Hardy's Course of Pure Mathematics?

We know that $3$ lengths can form a right angled triangle iff those lengths satisfy $a^2+b^2=c^2$.

But no matter what we choose for $m, \; n$ and $ \lambda$. $$(2\lambda mn)^2+(\lambda(m^2-n^2))^2=4\lambda^2m^2n^2+\lambda^2m^4-2 \lambda^2m^2n^2+\lambda^2n^4$$ $$=\lambda^2m^4+2\lambda^2m^2n^2+\lambda^2n^4=(\lambda(m^2+n^2))^2$$ as we wanted.

So we can choose any number of rationals $m, \; n$ and $\lambda$ to produce any number of right-angled triangles.

A simple example of an infinite solution set is if you keep $m$ and $n$ constant, but vary $\lambda$, you will get infinite solutions as there are an infinite amount of rational numbers.