Exercise 5.6.14 (Rational points on the curve $x^3 +y^3 = 7$)

Proof. To treat simultaneously Theorem 5.16 and Problem 14, we consider the cubic ${𝒞}_f(\mathbb{Q})$, where

( $c=9$ in Theorem 5.16, and $c=7$ in Problem 14.)

Suppose that $(a,b)\in {𝒞}_f(\mathbb{Q})$, so that $a^3+b^3=c$. Then

The tangent $T$ to the curve at point $(a,b)$ has equation

thus the equation of $T$ is

The point of intersection $(x,y)$ of the curve with the tangent (distinct of $(a,b)$) is given by the equations

The abscissa of the point of intersection is given by the equation

Multiplying by $b^6$, we obtain

Since $b^3=c-a^3$, this gives the equation

Dividing by $c\ne 0$,

Since $a$ is a double root, we may factorize ${(x-a)}^2$. This gives

where $x\ne a$, thus

and

so

Suppose that $(X_n,Y_n,Z_n)$ is an integer solution of $X^3+Y^3=c{Z}^3$ satisfying $Z_n\ne 0$, so that $(a,b)=(X_n∕Z_n,Y_n∕Z_n)\in {𝒞}_f(\mathbb{Q})$. Then (1) gives another rational point of the curve $(X_{n+1}∕Z_{n+1},Y_{n+1}∕Z_{n+1})$, such that

This gives another solution $(X_{n+1},Y_{n+1},Z_{n+1})$ of $X^3+Y^3=c{Z}^3$, defined by

The recurrences are those of Theorem 5.16, and are independent of $c\ne 0$.

With $c=7$, $(X_0,Y_0,Z_0)=(2,-1,1)$ is solution. Using (3), we obtain the new solutions

We must prove that there are infinitely many distinct rational points on the curve $x^3+y^3=7$. As in Theorem 5.16, since $X_0=2$, and $X_n\mid X_{n+1}$ for all $n$, we obtain

Since $Y_0$ and $Z_0$ are odd, we obtain by induction with formulas (3) that

Since $X_n^3\equiv 0(\mathit{mod}4)$, and $Y_n^3\equiv 1,-1(\mathit{mod}4)$, so $2Y_n^3\equiv 2(\mathit{mod}4)$, it follows that $X_n^3+2Y_n^3\equiv 2(\mathit{mod}4)$. Therefore ${\nu }_2(X_n^3+2Y_n^3)=1$, thus by (3)

Since ${\nu }_2(X_0)=1$, this shows that ${\nu }_2(X_n)=n+1$. Since $Y_n$ is always odd, ${\nu }_p(X_n∕Y_n)=n\ne {\nu }_p(X_m∕Y_m)=m$ if $n\ne m$, so $X_n∕Y_n\ne X_m∕Y_m$. We have proved that the curve $x^3+y^3=7$ has infinitely many distinct rational points. □