Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 125 / 5

Fix a prime $p$. If a rational point $T=(x,y)$ has $v_p(x)<0$, the integral projective Weierstrass equation forcesfor some $m>0$. Hence its formal parameter $t=-x/y$ lies in $p{\mathbb{Z}}_p$, so $T$ belongs to the formal group of an elliptic curve at the identity. This formal kernel has no nonzero torsion for the present equation. Multiplication by an integer prime to $p$ is a formal-group automorphism, while the formal logarithm rules out $p$-power torsion for odd $p$. For $p=2$, inversion sends $t$ to $-t$ because the equation has no $xy$ or $y$ term, and thereforeFor $v_2(t)\ge 1$, its leading term has strictly smaller valuation than every higher term, so $[2{]}_F(t)$ cannot vanish; iterating excludes all two-power torsion as well.

Thus a nonzero torsion point cannot have $v_p(x)<0$. Since this holds for every prime, $x\in \mathbb{Z}$. The integral equation then makes $y^2$ an integer; a rational number whose square is integral is itself integral, so $y\in \mathbb{Z}$. This is the integrality assertion in the Lutz–Nagell theorem.

For $E:y^2=x^3-x^2+25x$, the addition formulas giveIn particular, $[2]Q=T$ and $Q$ has order four, while the nonintegral coordinate of $[2]P=(144/25,-2172/125)$ and part (a) show that $P$ has infinite order.

For two-isogeny descent, writeThe rational point $(0,0)$ is the kernel of a degree-two isogeny $E\to E^{\prime }$. The square-class maps send an affine point with $x\ne 0$ to $[x]\in {\mathbb{Q}}^{\ast }/{\mathbb{Q}}^{\ast 2}$, send $O$ to $1$, and send $(0,0)$ to $[b]$ or $[a^2-4b]$ on the two curves. Their images determine the rank through

HereOn $E$, the possible square classes divide $25$; real solubility excludes the negative classes, while $P$ and $Q$ exhibit $1$ and $5$. Thus $\alpha (E(\mathbb{Q}))=\{1,5\}$. On $E^{\prime }$, the pointsexhibit the classes $-11,-1,11$, along with $1$. The remaining candidate classes are divisible by $3$. Their homogeneous spaceshave no primitive solution modulo $9$: reduction modulo $3$ first forces $3\mid UV$, and then the equation is congruent to $3$ or $6$ modulo $9$. HenceThe rank formula gives $2^r=2\cdot 4/4=2$, so $r=1$.

At the good primes $7$ and $19$, direct counts giveThe reduction of torsion points on an elliptic curve injects rational torsion into both groups, so its order divides $\gcd (12,16)=4$. Since $Q$ has order four, the torsion subgroup is $\mathbb{Z}/4\mathbb{Z}$. Thereforeso $t=1$, $d_1=4$, and $r=1$.