Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 125 / 2 / b

For $E:y^2=x^3+x+1$, the duplication formula isAt $P=(0,1)$, the tangent slope is $1/2$, so $x(2P)=1/4$ and $v_2(x(2P))=-2$. If $v_2(x(Q))=-2m<0$, the unique lowest-valuation terms in the numerator and denominator are respectively $x(Q{)}^4$ and $4x(Q{)}^3$, givingInduction yields $v_2(x(2^{m}P))=-2m$.

For a minimal integral equation, let $t=-x/y$ be the parameter of the formal group of an elliptic curve. Definewhere $E_1$ is the kernel of reduction to the identity. The parameter identifies $E_r$ with the formal group on $p^r{\mathbb{Z}}_p$. For odd $p$, the formal logarithm converges on $p{\mathbb{Z}}_p$ and is an analytic group isomorphism

For $p=2$, the logarithm gives $E_2({\mathbb{Q}}_2)\cong 4{\mathbb{Z}}_2$. The formal duplication series satisfies $[2](T)=2T+O(T^3)$, so $[2](2u)/4\equiv u(\mathrm{mod}2)$; successive lifting makes $[2]:E_1\to E_2$ surjective. Its kernel is rational 2-torsion, but $x^3+x+1$ has no root in ${\mathbb{Q}}_2$ because it has no root modulo $2$. Thus $[2]$ is an isomorphism and

Solved by gpt-5.6-sol high.