Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 125 / 5

Let a rational right triangle have legs $A,B$, hypotenuse $C$, and area $AB/2=D$. After interchanging the legs if necessary, setUsing $C^2=A^2+B^2$ and $AB=2D$ givesso $P=(x,y)$ lies on the congruent number elliptic curve $E_D:D{y}^2=x^3-x$. The triangle is nondegenerate, so $y\ne 0$ and $2P\ne O$.

Conversely the three displayed lengths in the question satisfyand their area isThus every such triangle is ${\Delta }_P$.

The change of variables $X=5x$, $Y=25y$ identifies $E_5$ withA full two-isogeny descent, equivalently the standard full 2-descent for a cubic with three rational roots, givesThe local conditions at $2$, $5$, and infinity leave exactly these eight square-class combinations; primes outside $\{2,5\}$ have even valuations and contribute none. Hence the quotient has dimension three. Since the rational 2-torsion has dimension two, the rank is one.

The Lutz–Nagell theorem on the integral model, or reduction at two good primes, excludes odd torsion and torsion of order greater than two. Thereforeand $(-4/5,6/25)$ generates the free part. This is the Mordell-Weil group of the congruent number curve for five.

For $x(P)=A/B^2$ in lowest terms, take the logarithmic naive height $h(P)=\log \max \{|A|,B^2\}$. Its required properties areandwith constants depending only on the curve. Define the canonical height of an elliptic curve byThe first bounded-error relation makes this a convergent telescoping correction to $h(P)$. Apply the second relation to $2^{r}P,2^{r}Q$, divide by $4^r$, and let $r\to \infty$ to obtainAlso $\hat{h}(2P)=4\hat{h}(P)$, and the parallelogram identity then gives $\hat{h}(nP)=n^2\hat{h}(P)$ for every integer $n$. Thus $\hat{h}$ is a quadratic form.

By part b, writeThe canonical height of an elliptic curve vanishes on torsion and satisfies $\hat{h}(mG+T)=m^2\hat{h}(G)$. If $\hat{h}(P)=\hat{h}(Q)\ne 0$, their nonzero integer coefficients therefore have equal squares, sofor some $T\in E_5[2]$.

Negation leaves the $x$-coordinate unchanged. Addition by the four 2-torsion points changes $x$ amongSubstitution in the three side formulas for ${\Delta }_P$, using $5y^2=x^3-x$, only changes signs and permutes the three values. Hence ${\Delta }_Q$ and ${\Delta }_P$ are the same right triangle up to reordering their sides.