Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 125 / 1 / c / Solution

The Lutz–Nagell theorem says that ifhas nonzero discriminant and $P=(x,y)\in E(\mathbb{Q})$ is a torsion point, then $x,y\in \mathbb{Z}$ and either $y=0$ or

For integrality, fix a prime $p$. If a rational point has nonintegral coordinates, its primitive projective coordinates reduce to $O$, so it belongs to the kernel of reduction. The parameter $t=-x/y$ identifies this kernel with the formal group of an elliptic curve over $p{\mathbb{Z}}_p$. The formal logarithm, with the standard separate first-step argument at $p=2$, shows that this group has no nonzero rational torsion. A rational torsion point therefore has nonnegative $p$-adic valuations in both coordinates for every prime $p$, hence integral coordinates.

Suppose now that $y\ne 0$. The point $2P$ is again a nonzero torsion point and hence integral. Its $x$-coordinate is $m^2-2x$, whereThus $m^2$ is an integer. A rational number whose square is integral is integral, so $2y\mid 3x^2+a$ and in particular $y^2\mid (3x^2+a{)}^2$. The curve equation and the identitythen prove $y^2\mid 4a^3+27b^2$. This is the divisibility proof in the Nagell–Lutz theorem.