By Fermat's last theorem, the equation $u^3+v^3=w^3$ has no solutions in positive integers $u,v,w$. Now consider the following variant : call $\rho(x)$ the distance between $x$ and the nearest integer, for any real number $x$ (thus $\rho(3)=0,\rho(3.2)=0.2,\rho(3.5)=0.5, \rho(3.9)=0.1$ etc).

An "approximate" version of the Fermat equation is to ask for $\rho({( u^3+v^3 )}^{\frac{1}{3}})$ to be arbitrarily small. A trivial way to achieve this is to make one of the variables very small compared to the other, say $v$ very small compared to $u$, so that ${( u^3+v^3 )}^{\frac{1}{3}}$ is very near to $u$.

It is therefore natural to ask if there is an absolute constant $C>0$ such that $\rho({( u^3+v^3 )}^{\frac{1}{3}})$ can be made arbitrarily small with $u,v$ positive and $u \leq C v, v \leq C u$ (so that neither of $u$ or $v$ dominates).

Can a (reasonably) explicit $(u_n,v_n)$ sequence be found, such that $\rho(u_n^3+v_n^3)$ tends to $0$ as $n$ goes to infinity and $u_n \leq C v_n, v_n \leq C u_n$ ? "Closed-form" formula would be the best, of course, but even a simple recursion would be nice.

  • $\begingroup$ Why $\rho(3)=3$? I think that $\rho(3)=0$. And do you actually mean $\rho(\sqrt[3]{u^3+v^3})$? Then just consider $u_n=n, v_n=1$. $\endgroup$ Jan 23, 2010 at 12:12
  • $\begingroup$ Yes, indeed. I'll correct that. $\endgroup$ Jan 23, 2010 at 12:14
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    $\begingroup$ Or if you want u_n and v_n tending to infinity, u_n=n^10 and v_n=n will do. It's hard for me to see whether this can be turned into an interesting question. $\endgroup$ Jan 23, 2010 at 12:57
  • $\begingroup$ I'll make one last try, and delete this question if it turns out to be trivial again. $\endgroup$ Jan 23, 2010 at 13:27
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    $\begingroup$ Let p/q be a close approximation to 2^(1/3), say p/q = 2^(1/3) + e with |e| < 1/(sqrt(5)*q^2); there are infinitely many of these by Hurwitz's theorem. Letting u=v=q, we have (u^3+v^3)^(1/3) = 2^(1/3) * q = p - qe, so the error qe is at most 1/(sqrt(5)*q) in magnitude, and as q goes to infinity the error goes to 0. $\endgroup$ Jan 23, 2010 at 16:12

2 Answers 2


The condition you want is very weak, and there are clearly many accidental solutions.

You can add severe restrictions and still find many solutions. For example, (as Steven Sivek pointed out) you can force $u_n = v_n$ and then by the theory of simple continued fractions, there are infinitely many $p_n/q_n$ so that

$|\frac{p_n}{q_n}-\sqrt[3]2|\lt q_n^{-2}.$

Then $|p_n - q_n \sqrt[3]2| < 1/q_n$ so $\rho(\sqrt[3]{(q_n^3+q_n^3)} )$ decreases to 0 rapidly.

This might not be viewed as explicit since the coefficients of the simple continued fraction for $2^{1/3}$ don't have a clear pattern, although you can define them recursively if you allow functions like $\lfloor1/\rho\rfloor$.

If you can find $x^3 + y^3 = z^3$ in numbers with known simple continued faction expansions, then you may be able to use this to construct closed form families of examples. For example,

$(5-\sqrt{6})^3 + 3\sqrt{6}^3 = (5+\sqrt{6})^3$.

Convergents of $\sqrt6$, $p_n/q_n$, satisfy $|\frac{p_n}{q_n} - \sqrt{6}| < 1/q_n^2$.

Then $(5q_n-p_n)^3+(3p_n)^3 = (5q_n+p_n)^3 + O(q_n) = (5q_n+p_n)^3 + o((5q_n+p_n)^2)$,

so $\rho(\sqrt[3]{(5q_n-p_n)^3+(3p_n)^3}) = o(1).$

For example, $\sqrt[3]{(5\times 881 - 2158)^3 + (3 \times 2158)^3} = 6552.99916...$

Since $\sqrt{6} = [2;2,4,2,4,2,4...]$ which is periodic, you can construct a closed form expression for the convergents $p_n/q_n$.


Elkies' paper "Rational points near curves and small nonzero |x^3-y^2| via lattice reduction" is relevant to your question: he discusses the efficient computation of solutions to |x^3 + y^3 - z^3| < M, and remarks that there are conjectured to be infinitely many of these "near-Fermat" triples. Indeed, his argument is very general and is not restricted to algebraic curves: he finds, for instance, that

2063^pi + 8093^pi = 8128^pi + .019....!

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    $\begingroup$ I saw this question and thought, "Oh, this is a good chance to mention this cool paper of Elkies," and then I saw this answer and thought, "Oh well, someone already mentioned it," and then I realized it was me. $\endgroup$
    – JSE
    Dec 22, 2010 at 1:54

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