# For what $N$ does $N=x^3+y^3+z^3-2xyz$ have infinite integer solutions?

Further to this question, A hard Cubic Diophantine equation I considered the more general equation $N=x^3+y^3+z^3-2xyz$.

As $x^3+y^3+z^3-2xyz$ is homogenous, any solution $(N_0,x_0,y_0,z_0)$ to $x^3+y^3+z^3-2xyz=N$ gives a set of solutions $(k^3N_0,kx_0,ky_0,kz_0)$.

In particular, any solution with positive $N$ gives a solution for $-N$ using $k=-1$, so it’s not necessary to further consider negative $N$

For example, $(N,x,y,z)=(13,-3,2,2)$ gives $(-13,3,-2,-2)$

Clearly, $(x,y,z)$ are interchangeable.

Using a small search up to $N=152$ I’ve found, with $a,b,k$ integer, just

$$(N,x,y,z)=(0,-a,0,a)$$

$$(N,x,y,z)=(19k^3,(-b-1)k,3k,(b-1)k)$$

Examples as $(N,x,y,z)$

$$(0,-45,0,45)$$ $$(19,-6,3,4)$$ $$(152,-12,6,8)$$

I’ve noticed other $N$ that seem possible candidates, but haven’t spotted the patterns. For example,

$$(9,-1575,583,1163)$$ $$(9,-944,522,545)$$ $$(9,-703,-198,838)$$ $$(9,-323,-187,457)$$ $$(9,-167,80,108)$$ $$(9,-162,86,97)$$ $$(9,-47,-34,72)$$ $$(9,-7,4,4)$$ $$(9,-2,1,2)$$ $$(9,0,1,2)$$ $$(9,1,2,2)$$

Other $N$ values that look interesting are $6,17,33,37,48,51,72,93,96,107,114,117,136$

My question:

Apart from $0$ and numbers of the form $19k^3$, for what $N$ does $N=x^3+y^3+z^3-2xyz$ have infinite integer solutions?

Update 5th March 2018

I’m also interested in:

values of $N$ where all solutions are known

values of $N$ where it can be shown that there are a finite number of solutions

values of $N$ where it can be shown that there no solutions.

• The equation $x^3+y^3+z^3-2xyz=9$ curiously has a lot of integer solutions. What were the search bounds used? – Tito Piezas III Mar 6 '18 at 13:00
• @TitoPiezasIII I looked in the range $-3000$ to $3000$. The other $N$ I note as interesting are much the same. Up to $N=152$, without counting them all, $117$ perhaps has the most with $21$ solutions. Please let me know if you wish me to include these, or other, results within my question. – Old Peter Mar 6 '18 at 15:28
• The case $N=9$ will do for now. The situation is reminiscent to the problem $x^3+y^3+z^3 =N$ where some (like $N=792$) have many integer solutions, but is not known if there is infinitely many. – Tito Piezas III Mar 6 '18 at 16:01
• I think the general equation is $x^3+y^3+z^3+Mxyz = N$ where my post is $M=0$ while yours is $M=-2$. It seems general results are not yet known. – Tito Piezas III Mar 6 '18 at 16:04

## 1 Answer

Regarding the above equation shown below:

$N=x^3+y^3+z^3-2xyz$

Seiji Tomita has given an identity on his web site:

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

Click on the link "Computational number theory" & then select article #264.

His identity is $x^3+y^3+z^3-n(xyz)=(n^3-27)$

where $(x,y,z)=(m,n-m,-3)$

For $n=2$ we have,

$(-m)^3+(m-2)^3+3^3-2(m)(2-m)(3)=19$

Since $m$ can take infinite values, then $N = 19$ can take on infinite integer solutions.

• Thank you for your interesting answer, which I’ve upvoted. I think my questions already shows $19k^3$ as an answer, but Seiji Tomita’s identity provides a solution to more general problem. – Old Peter Mar 6 '18 at 15:49