Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2007 / B / 8

The doubling slope at $P=(2,1)$ is $(3\cdot 2^2-1)/(2\cdot 1)=11/2$, givingThese denominators are invertible modulo $N$. To add $P$ again, the difference of first coordinates is $105/4-2=97/4$, so the new slope would be $-1083/194$. This requires inversion of a number divisible by 97. The elliptic curve method instead computesIt is a proper factor. The failure of the group-law division modulo the composite integer reveals different behavior modulo its prime factors.

Write $x=c/d$ in lowest terms and put $H=\max \{|c|,|d|\}\ge 1$. For $P\ne O,(0,0)$,Before reduction to lowest terms, both numerator and denominator are bounded in magnitude by $(1+|a|+|b|)H^2$. Cancellation can only decrease the height. Hence one may chooseThe duplication formula for $C$ isThe numerator has magnitude at most $(1+|b|{)}^2{H}^4$, and the denominator at most $4(1+|a|+|b|)H^4$. Thussatisfies the second bound. At the identity or any point whose relevant image is the identity, the image height is zero and the same nonnegative constants work directly. Thus the formulas cover every rational point, including the exceptional points of the displayed coordinate expressions.

If $p\equiv 1(\mathrm{mod}4)$, choose $i\in {\mathbb{F}}_p$ with $i^2=-1$. The cubic has the three distinct roots $0,im,-im$, so all four two-torsion points are rational over ${\mathbb{F}}_p$. They form a subgroup of order four, hence $4\mid \#E({\mathbb{F}}_p)$ by Lagrange's theorem.

If $p\equiv 3(\mathrm{mod}4)$, let $\chi$ be the Legendre symbol with $\chi (0)=0$. Since $f(x)=x(x^2+m^2)$ is odd and $\chi (-1)=-1$, the terms for $x$ and $-x$ cancel in ${\sum }_x\chi (f(x))$. Counting the possible second coordinates givesThus in both casesThe assumptions $p\ne 2$ and $m\ne 0$ ensure nonsingularity and keep the displayed two-torsion roots distinct when they exist.