nt.number theory - Variant of Fermat's last theorem

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 $lfloor1/rhorfloor$.

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,

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

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

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

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