2
$\begingroup$

I am asked to find constants $A$ and $B$ such that any solution of: $x^3+2y^3=m$ satisfies $\max\{|x|,|y|\} \leq A|m|^B$.

I am told to use the fact that: $ \left| \frac{p}{q} - \sqrt[3]{2} \right| \geq \frac{1}{4|q|^{2.45}} $ for all $(p,q) \in \mathbb{Z}\times \mathbb{Z} \setminus \{0\}$.

However, I don't know where to begin. Clearly, we have that $ \max\{|x|,|y|\} \leq A|m|^B = A|x^3+2y^3|^B \leq A(|x|^3+2|y|^3)^B$

But I have no idea how to continue.

I need to bound $\max \{|x|,|y|\}$ by some exponential function, and feel that I need to use that fact that the convergents $p_n/q_n$ of $\sqrt[3]{2}$ provide solutions to the equation above.

I have tried setting $q = y$ or $q=m$ but didn't see anything helpful from the algebra. Any advice for general strategy in approaching this problem, would be appreciated.

1 Answers 1

1

Replacing $y$ with $-y$ we have $m=x^3-2y^3=(x-ay)(x^2+axy+a^2y^2)$ where $a=\root3\of2$. Then $x^2+axy+a^2y^2\ge by^2$ for some easily-calculated $b$, and $|x-ay|\ge(1/4)|y|^{-1.45}$, so $m\ge(b/4)y^{.55}$.