# The "stubborn" solutions to sums of three cubes

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.

• You might be able to deduce the ratio going to infinity bit by choosing a $k$ that is a cubic non-residue for the first several primes that are $1 \mod 3$. I think then the constants in Heath-Brown'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. Feb 2, 2021 at 2:41
• I always get nervous with anonymous references like [1], in case of link rot. The referenced paper is Heath-Brown - The density of zeros of forms for which weak approximation fails. Dec 2, 2021 at 0:31

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 $$ with no solutions with $$|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 $$\limsup_{ x\to\infty} \frac{S(x)}{ x^{1/3} \sqrt{ \log \log x }} >0$$.

• One can improve on this, maybe to $x^{1/3} e^{ \sqrt{ \log \log x}}$, by counting only points in a region like $|x|,|y|,|z|<m$, $|x^3+y^3+z^3|<n$. Feb 2, 2021 at 14:15
• Excellent! My one nitpick is that "pigeon" doesn't have a d in it. Feb 3, 2021 at 5:30
• It seems to me that there is a typo in your $p^3+6p^2$ formula? Shouldn't the answer be $1 \bmod p-1$, since this is a homogenous equation? I think the right answer is $p^3+6p^2-6p$. Feb 3, 2021 at 14:42
• @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 $3^4$. Feb 3, 2021 at 14:46
• @RosieF This is explained, and the solutions are counted, in the last two pages of Heath-Brown'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. Feb 3, 2021 at 14:49