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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 $$\{0,\ldots,5000\}\setminus\{x^3+2y^3+3z^3:\ x,y,z\in\{-30000,\ldots,30000\}\}$$ only contain 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 $x^3+2y^3+3z^3=36$?

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

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 $$\{0,\ldots,5000\}\setminus\{x^3+2y^3+3z^3:\ x,y,z\in\{-30000,\ldots,30000\}\}$$ only contain 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 $x^3+2y^3+3z^3=36$?

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 $$\{0,\ldots,5000\}\setminus\{x^3+2y^3+3z^3:\ x,y,z\in\{-30000,\ldots,30000\}\}$$ only contain 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 $x^3+2y^3+3z^3=36$?

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 $$\{0,\ldots,5000\}\setminus\{x^3+2y^3+3z^3:\ x,y,z\in\{-30000,\ldots,30000\}\}$$ only contain four numbers: $36,\ 288,\ 2304,\ 4500.$ 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 $x^3+2y^3+3z^3=36$?

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 $$\{0,\ldots,5000\}\setminus\{x^3+2y^3+3z^3:\ x,y,z\in\{-30000,\ldots,30000\}\}$$ only contain 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 $x^3+2y^3+3z^3=36$?