The "stubborn" solutions to sums of three cubes

Question

It is conjectured (see [1]) that for any integer $k\not\equiv \pm 4\pmod 9$ there are infinitely many integer solutions to $$ a^3+b^3+c^3=k. $$ Numerical investigations of this conjecture show that for some $k$ solutions are easily found (for example, $510=101^3-100^3-31^3$), while other $k$ kind of "resist" the numerical search. Notable example is $$ (−80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3=42 $$ found by A. Booker and A. Sutherland in 2019. But is this phenomenon known to exist for infinitely many $k$? More precisely, let $$ s(k)=\min_{a^3+b^3+c^3=k}\max\{|a|,|b|,|c|\}. $$ and $S(x)=\max\limits_{\substack{k\leq x\\ k\not\equiv \pm 4\pmod 9}} s(k)$. Obviously, we don't even know if $S(114)$ is finite, so there is no upper bound for $S$ known. But what can be said about lower bounds for $S(x)$ for large $x$? Obviously, $S(x)\gg x^{1/3}$. Is it at least true that $$ \limsup_{x\to +\infty} \frac{S(x)}{x^{1/3}}=+\infty? $$ I think that a solid case for existence of "stubborn" integers would be a bound of the form $S(x)\gg x^{f(x)}$ with $f(x)\to +\infty$ as $x\to +\infty$ or at least $S(x)\gg x^A$ for some large $A$ (like $A=100$, for example), but I don't know if anything like this is within the reach of current techniques or even likely to be true.

Answer 1

This lim sup indeed goes to $\infty$. We can prove this using exactly the strategy Lucia suggested.

We will count the number of $x,y,z$ in a box with $x^3+y^3+z^3$ not a cubic residue modulo $p$ for a large finite list of primes $p$, all congruent to $1$ mod $3$. We can similarly count the number of $n$ not a cubic residue modulo $p$ for the same finite list of primes. From the pigeonhole principle, we can deduce that some $n$ is not a cubic residue for any $x,y,z$ in the box.

For $p$ congruent to $1$ mod $3$, the number of solutions of $x^3 + y^3 + z^3 + w^3=0 $ in $\mathbb F_p$ is $$ p^3 + 6 p^2-6p $$ and the number of solutions to $x^3 + y^3 + z^3 = 0$ is $$1 + (p-1 ) (p - a_p + 1) = p^2 - a_p (p-1) $$ where $|a_p| < 2 \sqrt{p}$ so the number of $x,y,z\in \mathbb F_p$ such that $x^3+y^3+z^3$ is a perfect cube is $$ \frac{ p^3 + 6 p^2 -6p + 2 ( p^2 - a_p (p-1) )}{3} $$ and thus the number of $x,y,z$ such that $x^3+y^3+z^3$ is not a perfect cube is

$$ p^3 - \frac{ p^3 + 6 p^2 -6p + 2 ( p^2 - a_p (p-1) )}{3} = \frac{ 2 p^3 - 8 p^2 - 2 a_p (p-1)}{3} .$$

Now let $c>0$ be an integer, and assume that $n>0$ is divisible by $\prod_{i=1}^k p_i^3$. Then the number of $x,y,z$ with $|x|, |y|, |z| < c n^{1/3} $ such that $x^3+y^3 + z^3$ is not a perfect cube mod any of $p_1,\dots p_k$ is

$$8 c^3 n \prod_{i=1}^{k} \frac{ 2 p_i^3 - 8 p_i^2 - 2 a_{p_i} (p_i-1) + 6p_i }{3p_i^3},$$

while the number of numbers less then $n$ that are not perfect cubes mod any of the $p_1,\dotsc, p_k$ is

$$ 2 n \prod_{i=1}^k \frac{ 2 (p_i-1)}{ 3 p_i }, $$

Therefore, by the pigeonhole principle, there must be some number $<n$ with no solutions with $|x|,|y|,|z| < cn^{1/3}$ as soon as

$$ 8 c^3 n \prod_{i=1}^{k} \frac{ 2 p_i^3 - 8 p_i^2 - 2 a_{p_i} (p_i-1) +6p_i}{3p_i^3}< 2 n \prod_{i=1}^k \frac{ 2 (p-1)}{ 3 p },$$ which happens when

$$ 4 c^3 < \prod_{i=1}^k \frac{ p_i^2 (p_i-1) }{ p_i^3 - 4 p_i^2 - a_{p_i} (p_i-1) + 3p_i} .$$

To check that we can take the integer $c>0$ sufficiently large, which lets the lim sup go to $\infty$, we need only ensure that the product goes to $\infty$ as the list $p_1,\dotsc, p_k$ grows to include all primes congruent to $1$ mod $3$. This is not hard to do as each factor is approximately $\frac{1}{ (1-p_i^{-1})^3}$ and the divergence follows from the fact that the primes congruent to $1$ mod $3$ have Dirichlet density $1/2 > 0$.

If we take $p_1,\dots, p_k$ the $k$ smallest primes congruent to $1$ mod $3$, and $n^{1/3}$ to be $\prod_{i=1}^k p_i$, I believe this method gives a lower bound like $\limsup_{ x\to\infty} \frac{S(x)}{ x^{1/3} \sqrt{ \log \log x }} >0$.