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

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

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  • $\begingroup$ Also known as $x(x+1)^3+y(y+1)^3+z(z+1)^3=x^4+y^4+z^4$. $\endgroup$ – Gerry Myerson Sep 5 '19 at 11:38
  • $\begingroup$ @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 ? $\endgroup$ – McRonald Sep 5 '19 at 13:29
  • $\begingroup$ According to MATLAB, not in $\{-100,\dots,100\}^3$. $\endgroup$ – Steve Huntsman Sep 5 '19 at 13:39
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    $\begingroup$ $(x + \frac{1}{3})^3 + (y + \frac{1}{3})^3 + (z + \frac{1}{3})^3 = \frac{1}{9}$ may also be useful. Or $(3x + 1)^3 + (3y + 1)^3 + (3z + 1)^3 = 3$. $\endgroup$ – user44191 Sep 5 '19 at 17:03
  • $\begingroup$ @SteveHuntsman According to the paper linked by user44191, any other solution must have $|x|,|y|,|z|>\frac1310^{16}$. $\endgroup$ – Emil Jeřábek Sep 5 '19 at 17:30
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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.

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    $\begingroup$ 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 $a\equiv b\equiv c\equiv1\pmod3$. $\endgroup$ – Emil Jeřábek Sep 5 '19 at 17:21

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