Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 125 / 3

On $y^2=x^3+5x^2+4x$, the slope through $P_1=(-1,0)$ and $P_2=(-2,2)$ is $-2$. The elliptic-curve addition formula givesFor doubling $P_2$, the tangent slope isand hence

The elliptic-curve discriminant is supported at $2$ and $3$, so $5$ and $7$ are primes of good reduction. Direct point counts giveFor example, summing $1+{\chi }_p(x(x+1)(x+4))$ over $x\in {\mathbb{F}}_p$ and adding the point at infinity gives these values.

The reduction of torsion points on an elliptic curve at the two primes shows that $|E(\mathbb{Q}{)}_{\mathrm{tors}}|$ divides eight. Part (a) shows that $P_2$ has order four, and $P_1$ is an independent point of order two because it does not lie in $⟨P_2⟩$. They already generate eight points, so

Use the two-descent on an elliptic curve map associated with the three rational roots $0,-1,-4$,with the standard limiting values at the 2-torsion. Only the square classes of $-1$, $2$ and $3$ can occur, because all other numerator and denominator valuations in the three factors $x,x+1,x+4$ are even. Checking solubility over $\mathbb{R}$, ${\mathbb{Q}}_2$ and ${\mathbb{Q}}_3$ leaves exactlyThese four classes are represented respectively by $O$, $P_1$, $P_2$ and $P_1+P_2$. Thus $|E(\mathbb{Q})/2E(\mathbb{Q})|=4$. Since part (b) givesthe quotient by doubling has order $2^{r+2}$. Therefore

Let the common difference of $r^2,s^2,1,t^2$ be $d$. ThenFor $x=-2s^2$ and $y=2rst$,so $(-2s^2,2rst)\in E(\mathbb{Q})$.

Part (c) says every rational point is one of the eight torsion points from part (b). Among their $x$-coordinates, the only negative value of the form $-2s^2$ with $s\ne 0$ is $-2$, arising from $P_2$ or $-P_2$. Thus $s^2=1$, and the four-term arithmetic progression has common difference zero. ConsequentlyClearing denominators gives Euler's corresponding result for integer squares.