Exercise 5.7.9 (Rational points on the elliptic curve $y^2 = x^3 + x$)

Proof. Let $(x,y)\in {\mathbb{Q}}^2$ be a solution of $y^2=x^3+x$. If $x=0$, then $y=0$. We suppose now that $x\ne 0$. Then $m=y∕x\in \mathbb{Q}$, and $m^2{x}^2=x^3+x.$ Since $x\ne 0$,

By completing the square, we obtain ${\left(x-\frac{m^2}{2}\right)}^2=\frac{m^4}{4}-1$, and by multiplying by $4$, this gives

Put $n=2x-m^2$. Then $n\in \mathbb{Q}$, and the discriminant $m^4-4=n^2$ is the square of a rational number. So

Since $(m,n)\in {\mathbb{Q}}^2$, there are integers $X,Y,Z$ such that $Z\ne 0$ and $m=X∕Z,n=Y∕Z$. Then

If we put $a=YZ,b=Z,c=X$, we obtain an integral solution of

This is equation (5.29). By the proof of Theorem 5.10, this Diophantine equation has no positive solution. Here $b=Z\ne 0$, therefore $c^4>0$, so $c\ne 0$. If $a=0$, then $Y=0$, thus $n=0$. Since $n=2x-m^2$, this gives $x=\frac{m^2}{2}$. By equation (1), we obtain

Then $m^2=2$. This is impossible because $m$ is a rational number. Therefore $a\ne 0$.

But then $a^2+4b^2=c^4$ has a solution in positive integers $|a|>0,|b|>0,|c|>0$, which is impossible by the proof of Theorem 5.10.

This shows that the only rational point of the elliptic curve associate to $f(x,y)=y^2-x^3-x$ is $(0,0)$. So the group $E_f(\mathbb{Q})$ has two elements, the points $(0,0)$ and the point at infinity $(0:1:0)$, and $E_f(\mathbb{Q})\simeq C_2$. □