Are all integers not congruent to 6 modulo 9 of the form $x^3+y^3+3z^2$?

Question

Are all integers not congruent to 6 modulo 9 of the form $x^3+y^3+3z^2$ for possibly negative integers $x,y,z$?

We have the identity $ (-t)^3+(t-1)^3+3 t^2=3t-1$.

The only congruence obstruction we found is 6 modulo 9.

Answer 1

This is one of the first questions of Waring type with "mixed powers" to be investigated. See the paper

H. Davenport, H. Heilbronn "On Waring's Problem: Two Cubes and One Square", Proc. London Math. Soc. 43, 1938, pp. 73–104.

where the authors use the circle method to show that almost all integers that are not 6 modulo 9 can be represented as $x^3+y^3+3z^2$. I believe the question of representability of all integers that are not 6 modulo 9 (or even all sufficiently large such integers) is still open.

Answer 2

We can also say the each $n \in \mathbb{Z}$ with $n \equiv 3 \pmod 9$ is representable as $x^3 + y^3 + 3z^2$. This is because $$(-t)^3 + (t-9)^3 + 3(3t -13)^2 = 9t - 222$$ and $-222 \equiv 3 \pmod 9$. So, along with the identity in the question we can represent each integer congruent to $2$, $3$, $5$, or $8 \pmod 9$. This leaves only $0$, $1$, $4$, and $7 \pmod 9$.

More generally we have $$(-t)^3 + (t - a^2)^3 + 3(at - b)^2 = 3a(a^3 - 2b)t + 3b^2 - a^6$$ which is congruent to $-a^6 \pmod 3$. This is either $0$ or $-1 \pmod 3$ depending on if $3$ divides $a$ or not. Taking $a = 3$ and $b = 13$ we obtain the first equation which takes care of $n \equiv 3 \pmod 9$. If we take $a = 3$ and $b = 12$ we get $27t - 297$. This gives us $n \equiv 0 \equiv -297 \pmod{27}$ which is some further progress. Hence, $0 \pmod 9$ can be resolved by finding $9 \pmod {27}$ and $18 \pmod{27}$.

Answer 3

We found another approach based on integral points on genus 0 curves.

Let $G(x,z,a_1,a_2,a_3)=(a_1 x+a_2)^3+(-a_1 x+a_3)^3+3 z^2$.

For fixed $n$ and $a_i$, $G=n$ is degree two genus 0 curve and it might have infinitely many integral points, which gives infinitely many solutions to the OP.

WolramAlpna on solve (10x-5)^3+(-10x+3)^3+3*y^2=25 over integers.

Related paper is Representation of an Integer as a Sum of Four Integer Cubes and Related Problems

Added 22 Dec 2020

We can solve $n \equiv 1 \pmod{3}$

The solutions of $(-x-1)^3+(x+2)^3 - 3y^2=(3N+1)$ are $x=-1/2\sqrt{4y^2 + 4N + 1} - 3/2$.

When we express $4N+1$ as difference of two squares $4N+1=z^2-y'^2$ with $y'$ even we have solution.

Currently the only unsolved case appears to be $n \equiv 0 \pmod{9}$.