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, $510=101^3100^331^3$), 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.

16$\begingroup$ You might be able to deduce the ratio going to infinity bit by choosing a $k$ that is a cubic nonresidue for the first several primes that are $1 \mod 3$. I think then the constants in HeathBrown's conjectured asymptotic would be small, and then a counting argument along such values of $k$ might do the job. Of course I haven't done the work. $\endgroup$– LuciaFeb 2, 2021 at 2:41

6$\begingroup$ I always get nervous with anonymous references like [1], in case of link rot. The referenced paper is HeathBrown  The density of zeros of forms for which weak approximation fails. $\endgroup$– LSpiceDec 2, 2021 at 0:31
1 Answer
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 $$ p^3 + 6 p^26p $$ and the number of solutions to $x^3 + y^3 + z^3 = 0$ is $$1 + (p1 ) (p  a_p + 1) = p^2  a_p (p1) $$ 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 $$ \frac{ p^3 + 6 p^2 6p + 2 ( p^2  a_p (p1) )}{3} $$ and thus the number of $x,y,z$ such that $x^3+y^3+z^3$ is not a perfect cube is
$$ p^3  \frac{ p^3 + 6 p^2 6p + 2 ( p^2  a_p (p1) )}{3} = \frac{ 2 p^3  8 p^2  2 a_p (p1)}{3} .$$
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
$$8 c^3 n \prod_{i=1}^{k} \frac{ 2 p_i^3  8 p_i^2  2 a_{p_i} (p_i1) + 6p_i }{3p_i^3},$$
while the number of numbers less then $n$ that are not perfect cubes mod any of the $p_1,\dotsc, p_k$ is
$$ 2 n \prod_{i=1}^k \frac{ 2 (p_i1)}{ 3 p_i }, $$
Therefore, by the pigeonhole principle, there must be some number $<n$ with no solutions with $x,y,z < cn^{1/3}$ as soon as
$$ 8 c^3 n \prod_{i=1}^{k} \frac{ 2 p_i^3  8 p_i^2  2 a_{p_i} (p_i1) +6p_i}{3p_i^3}< 2 n \prod_{i=1}^k \frac{ 2 (p1)}{ 3 p },$$ which happens when
$$ 4 c^3 < \prod_{i=1}^k \frac{ p_i^2 (p_i1) }{ p_i^3  4 p_i^2  a_{p_i} (p_i1) + 3p_i} .$$
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 $\frac{1}{ (1p_i^{1})^3}$ 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 $\limsup_{ x\to\infty} \frac{S(x)}{ x^{1/3} \sqrt{ \log \log x }} >0$.

2$\begingroup$ One can improve on this, maybe to $x^{1/3} e^{ \sqrt{ \log \log x}}$, by counting only points in a region like $x,y,z<m$, $x^3+y^3+z^3<n$. $\endgroup$ Feb 2, 2021 at 14:15

3$\begingroup$ Excellent! My one nitpick is that "pigeon" doesn't have a d in it. $\endgroup$– LuciaFeb 3, 2021 at 5:30

1$\begingroup$ It seems to me that there is a typo in your $p^3+6p^2$ formula? Shouldn't the answer be $1 \bmod p1$, since this is a homogenous equation? I think the right answer is $p^3+6p^26p$. $\endgroup$ Feb 3, 2021 at 14:42

1$\begingroup$ @RosieF In fact $n=2 \mod 7$ and $n=3 \mod 7$ have the same number of solutions  there are $27$ $(x,y,z)$ with $x^3=y^3=z^3=1\mod 7$, and also $27$ where two are $1$ and the other is $0$  the fewer number of choices that cube to $0$ exactly balances the choice of which one cubes to $0$. The correct thing is to multiply the the number of solutions mod $p$ divided by $p^2$ over all primes $p\neq 3$ times the number of solutions mod $9$ divided by $3^4$. $\endgroup$ Feb 3, 2021 at 14:46

1$\begingroup$ @RosieF This is explained, and the solutions are counted, in the last two pages of HeathBrown's paper The Density of Zeros of Forms for which Weak Approximation Fails. The specific densities for small $k$ were probably calculated along with the recent computer search efforts so you could look there. The "fake stubbon" examples you give are congruent to $8, 7,$ and $1$ mod $7$  probably the fact that they are not congruent to $\pm 3$ mod $9$ is the source of their fallure to be stubborn. $\endgroup$ Feb 3, 2021 at 14:49