Exercise 5.6.13 (Minoration of the solutions of $X_n^3 + Y_n^3 = 9Z_n^3$)

Proof.

(a)

By definition of the sequence ${(X_n,Y_n,Z_n)}_{n\in \mathbb{N}}$, $(X_0,Y_0,Z_0)=(2,1,1)$ and for all $n\in \mathbb{N}$, $$\begin{array}{llll}\hfill X_{n+1}& =X_n(X_n^3+2Y_n^3),& \hfill & \\ \hfill Y_{n+1}& =-Y_n(2X_n^3+Y_n^3),& \hfill & \\ \hfill Z_{n+1}& =Z_n(X_n^3-Y_n^3).& \hfill & \end{array}$$

Let $H_n=\max <!--\; FUNCTION\; APPLICATION\; -->(|X_n|,|Y_n|)$.

• If $|X_n|\ge |Y_n|$, then $H_n=|X_n|$.

  • If $|Y_n{\text{|}}^3\le \frac{1}{4}{H}_n^3$, then

    $$\begin{array}{llllll}\hfill |X_{n+1}|& =|X_n||X_n^3+2Y_n^3|& \hfill & & \hfill & \\ \hfill & \ge |X_n|(|X_n{\text{|}}^3-2|Y_n{\text{|}}^3)& \hfill (\text{by triangular inequality})& & \hfill & & \hfill \\ \hfill & \ge H_n\left(H_n^3-\frac{1}{2}{H}_n^3\right)& \hfill (\text{ since }{H}_n=|X_n|\text{ and }|Y_n{\text{|}}^3\le \frac{1}{4}{H}_n^3)& & \hfill & & \hfill \\ \hfill & =\frac{1}{2}{H}_n^4.& \hfill & & \hfill & \end{array}$$

    Then $H_{n+1}=\max <!--\; FUNCTION\; APPLICATION\; -->(|X_{n+1}|,|Y_{n+1}|)\ge \frac{1}{2}{H}_n^4$.

  • If $|Y_n{\text{|}}^3>\frac{1}{4}{H}_n^3$, then $|Y_n|\ge \frac{1}{\sqrt[3]{4}}{H}_n\ge \frac{1}{2}{H}_n$

    ( $\frac{1}{\sqrt[3]{4}}\ge \frac{1}{2}⟺\sqrt[3]{4}\le 2⟺4\le 2^3=8$) , so

    $$\begin{array}{llllll}\hfill |Y_{n+1}|& =|Y_n||2X_n^3+Y_n^3|& \hfill & & \hfill & \\ \hfill & \ge |Y_n|(2|X_n{\text{|}}^3-|Y_n{\text{|}}^3)& \hfill (\text{by triangular inequality})& & \hfill & & \hfill \\ \hfill & \ge \frac{1}{2}{H}_n(2|X_n{\text{|}}^3-|Y_n{\text{|}}^3)& \hfill (\text{since }|Y_n|>\frac{1}{2}{H}_n)& & \hfill & & \hfill \\ \hfill & \ge \frac{1}{2}{H}_n|X_n{\text{|}}^3& \hfill (\text{since }|Y_n|\le |X_n|)& & \hfill & & \hfill \\ \hfill & \ge \frac{1}{2}{H}_n^4& \hfill (\text{since }|X_n|=H_n).& & \hfill & & \hfill \end{array}$$

    Then $H_{n+1}=\max <!--\; FUNCTION\; APPLICATION\; -->(|X_{n+1}|,|Y_{n+1}|)\ge \frac{1}{2}{H}_n^4$.

• If $|X_n|\le |Y_n|$, then $H_n=|Y_n|$. Exchanging the roles of $X_n,Y_n$, we obtain similar inequalities.

  • If $|X_n{\text{|}}^3\le \frac{1}{4}{H}_n^3$, then

    $$\begin{array}{llllll}\hfill |Y_{n+1}|& =|Y_n||Y_n^3+2X_n^3|& \hfill & & \hfill & \\ \hfill & \ge |Y_n|(|Y_n{\text{|}}^3-2|X_n{\text{|}}^3)& \hfill (\text{by triangular inequality})& & \hfill & & \hfill \\ \hfill & \ge H_n\left(H_n^3-\frac{1}{2}{H}_n^3\right)& \hfill (\text{ since }{H}_n=|Y_n|\text{ and }|X_n{\text{|}}^3\le \frac{1}{4}{H}_n^3)& & \hfill & & \hfill \\ \hfill & =\frac{1}{2}{H}_n^4.& \hfill & & \hfill & \end{array}$$

    Then $H_{n+1}=\max <!--\; FUNCTION\; APPLICATION\; -->(|X_{n+1}|,|Y_{n+1}|)\ge \frac{1}{2}{H}_n^4$.

  • If $|X_n{\text{|}}^3>\frac{1}{4}{H}_n^3$, then $|X_n|\ge \frac{1}{\sqrt[3]{4}}{H}_n\ge \frac{1}{2}{H}_n$, so

    $$\begin{array}{llllll}\hfill |X_{n+1}|& =|X_n||2Y_n^3+X_n^3|& \hfill & & \hfill & \\ \hfill & \ge |X_n|(2|Y_n{\text{|}}^3-|X_n{\text{|}}^3)& \hfill (\text{by triangular inequality})& & \hfill & & \hfill \\ \hfill & \ge \frac{1}{2}{H}_n(2|Y_n{\text{|}}^3-|X_n{\text{|}}^3)& \hfill (\text{since }|X_n|>\frac{1}{4}{H}_n)& & \hfill & & \hfill \\ \hfill & \ge \frac{1}{2}{H}_n|Y_n{\text{|}}^3& \hfill (\text{since }|X_n|\le |Y_n|)& & \hfill & & \hfill \\ \hfill & \ge \frac{1}{2}{H}_n^4& \hfill (\text{since }|Y_n|=H_n).& & \hfill & & \hfill \end{array}$$

    Then $H_{n+1}=\max <!--\; FUNCTION\; APPLICATION\; -->(|X_{n+1}|,|Y_{n+1}|)\ge \frac{1}{2}{H}_n^4$.

In every case,

$$H_{n+1}\ge \frac{1}{2}{H}_n^4.$$

(b)

We cannot prove the proposed inequality by induction, but we will prove $$𝒫(n):H_n\ge \frac{1}{2^{\frac{4^{n-2}-1}{3}}}{H}_2^{4^{n-2}}$$

by induction for $n\ge 2$. Note that $\frac{4^{n-2}-1}{3}$ is an integer for any $n\ge 2$, because $4^{n-2}-1\equiv 0(\mathit{mod}3)$.

• $𝒫(2)$ is equivalent to $H_2\ge H_2$, so $𝒫(2)$ is true.

• Suppose that $𝒫(n)$ is true for some $n\ge 2$. Then, by part (a),

  $$\begin{array}{llll}\hfill H_{n+1}& \ge \frac{1}{2}{H}_n^4& \hfill & \\ \hfill & \ge \frac{1}{2}{\left(\frac{1}{2^{\frac{4^{n-2}-1}{3}}}{H}_2^{4^{n-2}}\right)}^4& \hfill & \\ \hfill & =\frac{1}{2^{\frac{4^{n-1}-4}{3}+1}}{H}_2^{4^{n-1}}& \hfill & \\ \hfill & =\frac{1}{2^{\frac{4^{n-1}-1}{3}}}{H}_2^{4^{n-1}}.& \hfill & \end{array}$$

  This shows that $𝒫(n+1)$ is true, if $𝒫(n)$ is true.

• This induction shows that

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->n\ge 2,H_n\ge \frac{1}{2^{\frac{4^{n-2}-1}{3}}}{H}_2^{4^{n-2}}.$$

Therefore

$$\begin{array}{llll}\hfill H_n& \ge \frac{1}{2^{\frac{4^{n-2}-1}{3}}}{H}_2^{4^{n-2}}& \hfill & \\ \hfill & \ge \frac{1}{2^{\frac{4^{n-2}}{3}}}{H}_2^{4^{n-2}}& \hfill & \\ \hfill & ={\left(\frac{H_2}{\sqrt[3]{2}}\right)}^{4^{n-2}}.& \hfill & \end{array}$$

By Problem 11, $H_2=188479>10\sqrt[3]{2}$, so

$$H_n\ge 10^{4^{n-2}}(n\ge 2).$$