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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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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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  • $\begingroup$ 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? $\endgroup$ – Amir Hossein Apr 13 '11 at 10:42
  • $\begingroup$ @Amir: for $n=2$ yes; it is the only known form but there are solutions that do not belong to any form. $\endgroup$ – Eelvex Apr 13 '11 at 10:47
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    $\begingroup$ An example which is not part of a known family is $1214928^3+3480205^3-3528875^3 = 2$. See euler.free.fr/identities.htm $\endgroup$ – Tito Piezas III Nov 25 '14 at 23:29
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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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  • $\begingroup$ Yes, the second one has infinitely many solutions. $\endgroup$ – Amir Hossein Apr 12 '11 at 16:02
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For \begin{align} x^3+y^3+z^3&=1\tag{1} \end{align} there are at least these one-parameter families of solutions \begin{gather} (9t^4)^3 + (-9t^4 \mp 3t)^3 + (\pm 9t^3 + 1)^3 = 1\tag{2}\\ (3888t^{10} - 135t^4)^3 + (-3888t^{10} - 1296t^7 - 81t^4 + 3t)^3 + (3888t^9 + 648t^6 - 9t^3 + 1)^3 = 1\\ \implies (27tu(144u^2-5))^3 + (-3t(27u(16u(3u+1)+1)-1))^3 + (9u(72u(6u+1)-1) + 1)^3 = 1, u=t^3\tag{3}\\ (-1679616a^{16}-559872a^{13}-27216a^{10}+3888a^7+63a^4-3a)^3 +\\ + (1679616a^{16}-66096a^{10}+153a^4)^3 +\\ + (1679616a^{15}+279936a^{12}-11664a^9-648a^6+9a^3+1)^3 = 1\\ \implies (3(-559872u^5-186624u^4-9072u^3+1296u^2+21u-1)a)^3 +\\ + (9(186624u^4-7344u^2+17)au)^3 +\\ + (1679616u^5+279936u^4-11664u^3-648u^2+9u+1)^3 = 1, u=a^3\tag{4} \end{gather} Beck attributes $(2)$ to Mahler (1936). For $(3)$ see mathpages. $(4)$ is the result of setting $b=1$ in an identity due to Kohmoto, cited by Wolfram, "Algebraic Identity".

Solutions to $(1)$ are a special case of "Fermat near-misses". Some are given in OEIS: \begin{align} A050787_i^3-A050789_i^3-A050788_i^3&=1\\ -A050791_i^3+A050793_i^3+A050792_i^3&=1 \end{align}

References:

Michael Beck et al. New integer representations as the sum of three cubes. Mathematics of Computation, 76, no.259 (Jul 2007), p.1683-1690. S 0025-5718(07)01947-3

Kurt Mahler. Note On Hypothesis K of Hardy and Littlewood. Journal of the London Mathematical Society, 11 (1936), p.136–138.

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