Let $m,n\in\mathbb{Z}$. Show that the curve $Y^2=X^3-(m^2+1)^2X+9n^2$ has infinitely many $\mathbb{Q}$-rational points.
This question was set in the context of an elliptic curves course. If we could find some 'parametrisation' $(X,Y)=(f(m),g(n))$ we would be done, but I suspect the elliptic curves idea would be useful here to find such functions. I suppose one idea is that this equation is in Weirstrass form.
An equation is in Weirstrass form if we have $y^2=x^3+Ax+B$, with discriminant $\Delta:=4A^3-27B^2$. In this case we have $\Delta=4((m^2+1)^2)^3+27(9n^2)^2$. I cannot see how to use this to find an infinite family of solutions though. Any help would be appreciated.
As in comment, I'd assume $n \neq 0$. There are some rational points on the curve (for any $m$ and $n$): $(0, \pm 3n)$ and $(\pm (m^2 + 1), \pm 3n)$. Now, to show that the curve as infinitely many rational points (i.e. rank is at least one), it is enough to show that at least one of these points is not a torsion. By Nagell-Lutz theorem, it is enough to show that $3n$ does not divide the discriminant, or equivalently, $3n$ does not divide $4(m^2 + 1)^6$. Now, it is an elementary number theory problem: can $m^2 + 1$ be a multiple of $3$?