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

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 $$x^3+2y^3+3z^3=36$$?

• I don't know if Andrew Booker's algorithm can be generalized from the sum of three cubes problem to this one, but surely Noam Elkies' algorithm can be adapted. May 11 '19 at 2:03
• oeis.org/A014136 May 11 '19 at 3:10
• @Bullet51 That is peculiar. The OEIS sequence lists 288 as a number that is not expressible in this form ...
– EGME
May 11 '19 at 14:21
• I don't think that your conjecture is any easier than the analogous conjecture on $x^3+y^3+z^3$. May 11 '19 at 14:33
• On my request, Andrew Booker obtained that the equation $x^3+2y^3+3z^3=36$ has no integral solutions with $\max\{|x|,|y|\}\le 10^8$. May 16 '19 at 19:12

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)

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

• But the question asks for integer solutions for every $n$, not rational solutions. May 16 '19 at 16:35
• There are "smaller" rational solutions like $(x,y,z)=(1/3, 10/3, -7/3)$ or $=(11/3, 2/3, -5/3 )$. Jun 16 '19 at 9:26