It is conjectured (see [[1](https://www.ams.org/journals/mcom/1992-59-200/S0025-5718-1992-1146835-5/home.html)]) 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.



The "stubborn" solutions to sums of three cubes
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 $\limsup\_{ x\to\infty} \frac{S(x)}{ x^{1/3} \sqrt{ \log \log x }} >0$.


