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In the book 'A Classical Approach to Modern Number Theory' by Ireland and Rosen, the penultimate paragraph of Chapter 17(Diophantine Equations) mentions,

Thus while $x^2 - 2y^2 = 1$ has infinitely many integral solutions, $x^3 -2y^3 = 1$ has only finitely many integral solutions.

I want to prove the latter fact. The polynomial $t^3 -2$ has a root at $\sqrt[3]{2}$.

I think I need to show that $$ \left| \sqrt[3]{2} - \frac{p}{q} \right| > \frac{c}{q^{3- \epsilon}} $$ for some $c, 3 > \epsilon > 0$ and apply the result proved in the previous paragraph which says

This is equivalent to showing that for some epsilon $|\sqrt[3]{2} - \frac{p}{q}| < \frac{1}{q^{3- \epsilon}}$ is satisfied by only finitely many rationals.

But I cannot prove this.

Subham Jaiswal
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    @coffeemath corrected now. – Subham Jaiswal Apr 15 '23 at 01:40
  • Some inequalities of the type you mention follow from the continued fraction expansion of the irrational approximated. [I don't know whether $\sqrt[3]{2}$ has a simple expansion.] – coffeemath Apr 15 '23 at 02:02
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    No, @coffeemath , the continued fraction expansion of $\sqrt[3]2$ looks very random, no rhyme or reason to it. – Lubin Apr 15 '23 at 02:29
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    Replace $y$ by $-y$ to arrive at $x^3 + 2 y^3 = 1.$ Theorem 5 on page 220 of Mordell, Diophantine Equations, says there is at most one integr solution with $xy \neq 0$ The proof continues to page 225. – Will Jagy Apr 15 '23 at 03:33
  • What are sample integer solutions for the first equation? Why do you say they are infinite? – NoChance Apr 15 '23 at 04:03
  • @NoChance The first equation $x^2-2y^2=1$ is called a Pell equation. It has infinitely many integers solutions, see for example this post. – Dietrich Burde Apr 15 '23 at 12:32

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