Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 22 / 3

In this one-dimensional setting, a commutative formal group law over a commutative ring $R$ is a formal power series $F(X,Y)\in R[[X,Y]]$ withIn particular $F(X,Y)=X+Y+$ terms of total degree at least two. There is a unique formal inverse $i_F(T)=-T+O(T^2)$; coefficient recursion solves $F(T,i_F(T))=0$.

A morphism from formal group law $F$ to $G$ is $f(T)\in TR[[T]]$ satisfyingThe invertible morphism criterion for formal group laws isNecessity follows by differentiating $g\circ f=T$ at zero for an inverse $g$. Conversely, write $f(T)=a_{1}T+a_2{T}^2+\cdots$ with $a_1$ a unit. In constructing $g(T)=b_{1}T+b_2{T}^2+\cdots$, the coefficient of $T$ fixes $b_1=a_1^{-1}$; at degree $n$, the equation $f(g(T))=T$ has the form $a_1{b}_n+$ an already known expression $=0$. This determines every $b_n$ over $R$. The same construction gives an inverse on the other side, and uniqueness makes the two inverses agree. Finally apply $g$ to the morphism identity with $X=g(U),Y=g(V)$ to obtainThus the inverse is itself a morphism of formal group laws, not merely an inverse formal power series. Over a general ring, nonzero derivative is insufficient: it must be a unit.

We use two precise facts about the formal group of an elliptic curve. At a prime $p$ of good reduction, its kernel of reduction of an elliptic curve is identified by the uniformizer $t=-x/y$ with the group $\hat{E}(p{\mathbb{Z}}_p)$. For odd $p$ this group is torsion-free. To see the second fact, the integral invariant differential of a formal group law has the form $(1+{\sum }_{j\ge 1}{c}_j{T}^j)dT$, with $c_j\in {\mathbb{Z}}_p$. Integrating constructs the formal logarithmIt is a group homomorphism to the additive group. For $0\ne t\in p{\mathbb{Z}}_p$ and $j\ge 2$,when $p$ is odd. Hence the series converges and $v_p({\log }_F(t))=v_p(t)$, so it is injective. The target has no nonzero torsion. Therefore reduction is injective on the entire rational torsion subgroup at an odd prime of good reduction, including its $p$-primary part.

For the present elliptic curve, $\Delta =64D^6$, so every odd $p\nmid D$ is a prime of good reduction. If $p\equiv 3(\mathrm{mod}4)$, the Legendre symbol of $-1$ is $-1$. The values $x$ and $-x$ cancel in the sum of the Legendre symbols of $x^3-D^{2}x$, yieldingThe rational torsion subgroup injects into each of these groups, so it is finite and its order $M$ divides every such $p+1$.

For any odd prime $\ell$, the Dirichlet theorem on primes in arithmetic progressions supplies infinitely many $p$ with $p\equiv 3(\mathrm{mod}4)$ and $p\equiv 1(\mathrm{mod}\ell )$. Discard the finitely many dividing $D$. Since $\ell \nmid p+1$, it cannot divide $M$. Similarly choose $p\equiv 3(\mathrm{mod}8)$, again avoiding $D$; then $v_2(p+1)=2$, so $M\mid 4$.

There are already four rational 2-torsion points,which are distinct because a squarefree integer $D$ is nonzero. ThusIts order is four. In fact the argument works for every nonzero integer $D$; squarefreeness is not needed for this torsion conclusion.

Use the congruent number elliptic curve in the equivalent coordinatesThe rational point $P=(25,100)$ lies on it, since $25^3-225\cdot 25=10000$. By the preceding rational torsion of a congruent number curve result, every rational torsion point of an elliptic curve has $y=0$ or is $O$. Thus $P$ is not a torsion point of an elliptic curve, and its positive multiples give infinitely many distinct rational points with nonzero $y$.

For any such point $(x,y)$, putThe identity $(x^2-225{)}^2+(30x{)}^2=(x^2+225{)}^2$ proves that these positive rational numbers are the sides of a right triangle. Their area isAll three sides are nonzero because a point with $y\ne 0$ has $x\notin \{0,15,-15\}$. For $P$ the construction gives $(a,b,c)=(4,15/2,17/2)$.

It remains to ensure that infinitely many points do not describe only finitely many triangles. Given the ordered positive pair $b,c$, set $r=c/b$. Then $X=|x|$ satisfiesThere are at most two possible $X$, then at most two signs of $x$ and two signs of $y$. Thus each ordered triangle has at most eight preimages; allowing interchange of its legs still gives a finite number. There are infinitely many distinct rational right triangles of area $15$. This is the infinitely many rational right triangles from a nontorsion point principle.