Exercise 5.6.15* (Rational points of the curve $x^3 + y^3 = 9$ in the first quadrant)

Proof. Let the triple $(X_n,Y_n,Z_n)$ of integers be defined by $(X_0,Y_0,Z_0)=(2,1,1)$ and

Then $X_n^3+Y_n^3=9Z_n^3$ (see Theorem 5.16 and Problem 14). Since $Z_n$ is odd by the proof of Theorem 5.16, $Z_n\ne 0$ for every $n\in \mathbb{N}$.

Put $x_n=X_n∕Z_n,y_n=Y_n∕Z_n$ for all $n\in \mathbb{N}$. Then $(x_n,y_n)$ is a rational point on the curve $x^3+y^3=9$, and

Note that $x_n=0$ is impossible, otherwise there is a first $n$ for which $x_n=0$. Since $x_0=2$, it follows that $n>0$. Then $X_n=0$, but $X_{n-1}\ne 0$, and $0=X_n=X_{n-1}(X_{n-1}^3+2Y_{n-1}^3)$, so $X_{n-1}^3+2Y_{n-1}^3=0$, which gives $Y_{n-1}∕X_{n-1}=-1∕\sqrt[3]{2}\in \mathbb{Q}$. But $\sqrt[3]{2}$ is an irrational number. This contradiction proves that $x_n\ne 0$ for all $n$, and similarly $y_n\ne 0$.

Since $x_n^3+y_n^3=9$, $x_n$ and $y_n$ cannot be both negative, therefore $(x_n,y_n)$ lie in the first quadrant if and only if $y_n∕x_n>0$. Put $m_n=y_n∕x_n$. By (2),

so, $m_0=1∕2$, and for all $n\in \mathbb{N}$, $m_{n+1}=h(m_n)$, where

We are led to study the recurrent sequence ${(m_n)}_{n\in \mathbb{N}}$, and to prove that $m_n>0$ for infinity many indices $n$. For all $x\in ℝ\setminus \{a\}$, where $a=-1∕\sqrt[3]{2}$,

and $h^{\prime }(x)=0⟺x=1$. Therefore $h$ is strictly decreasing on $]-\infty ,a[$ and on $]a,+\infty [$. Moreover

(see figure).

Since $h(x)=x⟺x=0$ or $x=-1$, $h$ has two fixed points $0$ and $-1$, which are repulsive because $h^{\prime }(0)=-2$ and $h^{\prime }(-1)=-8$ (this explains that the sequence ${(m_n)}_{n\in \mathbb{N}}$ is very chaotic).

Note that for every $n$, $m_n\ne 0$, because $y_n\ne 0$. Let $n$ be any positive integer. We want to prove that there is an index $n+k\ge n$ such that $m_{n+k}>0$. For this purpose, assume for the sake of contradiction that for all $k\ge 0$, $m_{n+k}<0$.

Since $m_{n+k}<0$ and $f(m_{n+k})=m_{n+k+1}<0$, we obtain using (4):

But $h$ is defined and decreasing on $I$. We prove by induction on $k$ that the sequence ${(m_{n+2k})}_{k\in \mathbb{N}}$ is monotonic (note that $(h\circ h)(m_{n+2k})=m_{n+2k+2}$).

Indeed, if $m_n\le m_{n+2}$, then the proposition $𝒫(k):m_{n+2k}\le m_{n+2k+2}$ is true for $k=0$. If $𝒫(k)$ is true for some $k\in \mathbb{N}$, then

so $𝒫(k+1)$ is true. This shows that for all $n\in \mathbb{N}$, $m_{n+2k}\le m_{n+2(k+1)}$, so the sequence ${(m_{n+2k})}_{k\in \mathbb{N}}$ is increasing.

Similarly, if $m_n\le m_{n+2}$, then ${(m_{n+2k})}_{k\in \mathbb{N}}$ is decreasing. In both cases, the sequence ${(m_{n+2k})}_{k\in \mathbb{N}}$ is monotonic.

The sequence ${(m_{n+2k})}_{k\in \mathbb{N}}$ is monotonic, and bounded (because $m_{n+k}\in [b,a[$), therefore convergent. Let $l=\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{m}_{n+2k}$. Then $l\in [b,a]$.

• If $l=a$, then $\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}|h(m_{n+2k})|=\infty$, thus $\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}|(h\circ h)(m_{n+2k})|=0$, so $\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{m}_{n+2k+2}=0$. Since $(m_{n+2k+2})$ is a sequence extracted from $(m_{n+2k})$, they have same limit, thus $a=0$. This is false, therefore $l\ne a$.

• Since $l\ne a$, then $l\in [a,b[$, thus $h$ is continuous at the point $l$, therefore

  $$(h\circ h)(l)=\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}(h\circ h)(m_{n+2k})=\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{m}_{n+2k+2}=l,$$

  so $l$ is a fixed point of $h\circ h$.

  $$\begin{array}{llll}\hfill 0& =(h\circ h)(x)-x=& \hfill & \\ \hfill & -\frac{3\left(x^4+2x^3-x+1\right)\left(x^4-x^3+3x^2-x+1\right)\left(x^4-x^3+2x+1\right)\left(x^2-x+1\right)\left(x+1\right)x}{\left(2x^{12}+4x^9+12x^6+10x^3-1\right)\left(2x^3+1\right)}& \hfill & \\ \hfill & ⟺x=0\text{ or }x=-1,& \hfill & \end{array}$$

  since none of the polynomials $x^4+2x^3-x+1,x^4-x^3+3x^2-x+1,x^4-x^3+2x+1,x^2-x+1$ has a real solution. So the fixed points of $h\circ h$ are the fixed points of $h$.

Since $0\notin [b,a]$, it remains only the possibility

This would be possible if $m_{n+2k}=-1$ for some $k$. But if $m_p=-1$ for some index $p$ then $y_p=-x_p$. This is impossible, because $9=x_p^3+y_p^3\ne 0$. Therefore for all $k\in \mathbb{N}$,

The contradiction follows from $|h^{\prime }(-1)|=8>1$, thus ${(h\circ h)}^{\prime }(-1)=h^{\prime }(h(-1))h^{\prime }(-1)={(h^{\prime }(-1))}^2=64>1$. Since $h$ is continuous at the point $-1$, there is some $\alpha >0$ such that $[-1-\alpha ,-1+\alpha ]\subset ]b,a[$ and

Since $\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{m}_{n+2k}=-1$, there is an integer $N>0$ such that

By the Mean Value Theorem, there is some $c\in [-1-\alpha ,-1+\alpha ]$ such that

therefore

By induction, this implies

We have proved that $m_{n+2N}\ne -1$, thus $\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{2}^k|m_{n+2N}+1|=+\infty$, so $\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}|m_{n+2N+2k}+1|=+\infty$, but $\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}|m_{n+2N+2k}+1|=0$. This is a contradiction, which proves that

Therefore infinitely many of the rational points $(X_n∕Z_n,Y_n∕Z_n)$ lie in the first quadrant. □