Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 125 / 5 / b / ii / Solution

For this curve, the square-class image in the first two-isogeny descent is contained in $\{1,-1,11,-11\}$. All four classes occur: $Q=(-1,1)$ gives $-1$, $T=(0,0)$ gives $11$, and $Q+T$ gives $-11$. The isogenous curve isIts image is contained in $\{1,-1,5,-5\}$. Negative $x$ cannot occur because $x(x^2-26x+125)<0$ for $x<0$, while $1$ and $5$ are represented by the identity and $(0,0)$. Thus the two image orders are four and two, andgives $r=1$.

The point $P$ has order three because $2P=-P$, and $T$ has order two, so the rational torsion contains a cyclic subgroup of order six. At the good primes $7$ and $13$, direct point counting givesReduction bounds the rational torsion order by their greatest common divisor, namely six, so this is all the torsion. The structure theorem for finitely generated modules over a principal ideal domain now givesThus one may take $d_1=1$, $d_2=6$, and $r=1$.

Solved by gpt-5.6-sol high.