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How do you show that the equation $x^3+y^3+z^3=1$ has infinitely many solutions in integers? How about $x^3+y^3+z^3=2$?

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2 Answers 2

up vote 13 down vote accepted

You can reduce the first equation to $$x^3 = -y^3, z = 1$$ with obvious infinite solutions. This paper details other families of solutions.

The second equation has solutions $(x,y,z)\equiv (6t^3+1, 1-6t^3, -6t^2)$ which (AFAIK) you find by construction (i.e you have to guess it).

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Yes, the solution you mentioned for the second one is exactly the same as the one I knew. Is this the only form of the solutions? –  Amir Hossein Apr 13 '11 at 10:42
@Amir: for $n=2$ yes; it is the only known form but there are solutions that do not belong to any form. –  Eelvex Apr 13 '11 at 10:47
An example which is not part of a known family is $1214928^3+3480205^3-3528875^3 = 2$. See euler.free.fr/identities.htm –  Tito Piezas III yesterday

For $x^3+y^3+z^3=1$ it is trivial - an infinite family of solutions is $(1,n,-n)$, and permutations of that.

For $x^3+y^3+z^3=2$ I'm not so sure there are infinitely many solutions. Are you just hypothesizing this, or do you know it to be true?

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Yes, the second one has infinitely many solutions. –  Amir Hossein Apr 12 '11 at 16:02

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