# What are the solutions of this Diophantine equation?

Besides $$(x, y, z)=(0, 0, 0)$$ and $$(1, 1, -2)$$ (and their permutations) are there any other integer solutions to the equation

$$3(x^{3}+y^{3}+z^{3})+3(x^{2}+y^{2}+z^{2})+(x+y+z)=0$$ ?

• Also known as $x(x+1)^3+y(y+1)^3+z(z+1)^3=x^4+y^4+z^4$. Sep 5, 2019 at 11:38
• @GerryMyerson, by ''also known as'', are you saying the question is a known open problem that can be expressed in the form that you wrote ? If yes, what is the name of the conjecture ? Any references ? Sep 5, 2019 at 13:29
• According to MATLAB, not in $\{-100,\dots,100\}^3$. Sep 5, 2019 at 13:39
• $(x + \frac{1}{3})^3 + (y + \frac{1}{3})^3 + (z + \frac{1}{3})^3 = \frac{1}{9}$ may also be useful. Or $(3x + 1)^3 + (3y + 1)^3 + (3z + 1)^3 = 3$. Sep 5, 2019 at 17:03
• @SteveHuntsman According to the paper linked by user44191, any other solution must have $|x|,|y|,|z|>\frac1310^{16}$. Sep 5, 2019 at 17:30

A simple transformation renders this equivalent to $$(3x + 1)^3 + (3y + 1)^3 + (3z + 1)^3 = 3$$; any solutions would give solutions to $$a^3 + b^3 + c^3 = 3$$ (and vice versa, as pointed out by Emil Jeřábek in a comment). According to a recent arxiv article, the only known solutions to the latter are $$(1, 1, 1), (4, 4, -5)$$ and its permutations, but the existence of other solutions remains open, so this question is also open.
Edited to add: This question seems to have been asked at a very opportune time; Booker and Sutherland have just found another solution: $$569936821221962380720^3−569936821113563493509^3−472715493453327032^3 = 3$$. That corresponds to $$x = 189978940407320793573, y = -189978940371187831170, z = -157571831151109011$$ for your question.
• Indeed, if $x\equiv0,\pm1\pmod3$, then $x^3\equiv0,\pm1\pmod9$, thus any solution to $a^3+b^3+c^3=3$ must have $a\equiv b\equiv c\equiv1\pmod3$. Sep 5, 2019 at 17:21