Is it true that $\{x^3+2y^3+3z^3:\ x,y,z\in\mathbb Z\}=\mathbb Z$?

Question

It is easy to see that no integer congruent to $4$ or $-4$ modulo $9$ can be written as the sum of three integer cubes. In view of this and Question 331163, I proposed the following conjecture in March 2019.

Conjecture. Every integer $n$ can be written as $x^3+2y^3+3z^3$ with $x,y,z$ integers. That is, $$\{x^3+2y^3+3z^3:\ x,y,z\in\mathbb Z\}=\mathbb Z.$$

This conjecture has an interesting application. Under the conjecture, my result on Hilbert's Tenth Problem implies that there is no effective algorithm to test for a general polynomial $P(x_1,\ldots,x_{33})$ with integer coefficients whether the diophantine equation $$P(x_1^{3},\ldots,x_{33}^3)=0$$ has integer solutions.

Quite recently, my PhD student Chen Wang checked my above conjecture seriously. He found that the set $$\{0,\ldots,5000\}\setminus\{x^3+2y^3+3z^3:\ x,y,z\in\{-30000,\ldots,30000\}\}$$ only contains four numbers: $36,\ 288,\ 2304,\ 4500.$ For example, he obtained that $$3772=(-20027)^3+2\times15936^3+3\times(-2739)^3.$$ Note that $$288=2^3\times 36,\ \ 2304=4^3\times36,\ \ 4500=5^3\times36.$$ So, to finish the verification of the conjecture for all $n=0,\ldots,5000$, it remains to find $x,y,z\in\mathbb Z$ with $x^3+2y^3+3z^3=36$.

QUESTION. Are there integers $x,y,z$ satisfying $x^3+2y^3+3z^3=36$?

Answer 1

Above equation shown below:

$x^3+2y^3+3z^3=n$ ------------$(1)$

Above equation $(1)$ can be written as:

$ax^3+by^3+cz^3=n$

Seiji Tomita has shown that for $(a+b=c)$ there are rational solution's for any '$n$'.

For, $n=36$, $(a,b,c)=(1,2,3)$, the solution is:

$(x,y,z)=[(167/9),(158/9),(-161/9)]$

The above numerical solution is equivalent to:

$(167)^3+2(158)^3+3(-161)^3=26244$

And, 26244= (9)^3*(36)

His web page link is:

     http://www.maroon.dti.ne.jp/fermat

Click on "Computational number theory" & select article # 313.