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For fixed $n \in \mathbb{N}$ consider integer solutions to $$x^3+y^3+z^3=n \qquad (1) $$

If $n$ is a cube or twice a cube, identities exist.

Elkies suggests no other polynomial identities are known.

For which $n$ (1) has infinitely many integer solutions?

Added

Is there $n$, not a cube or twice a cube, which allows infinitely many solutions?

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    $\begingroup$ Let $n = 1, 2$ and we have solutions for $x, y, z$ with a value $t \in \mathbb{N}$ however not all $x, y, z$ are positive, though I think this may serve as a little bit of help: $$\begin{align} 1 &= (9t^3 + 1)^3 + (9t^4)^3 + (-9t^4 - 3t)^3 \\ 2 &= (6t^3 + 1)^3 + (-6t^3 - 1)^3 + (-6t^2)^3 \end{align}$$ or for big solutions for $n = 1$: $$1 = (1 - 9t^3 + 648t^6 + 3888t^9)^3 + (-135t^4 + 3888t^{10})^3 + (3t - 81t^4 - 1296t^7 - 3888t^{10})^3$$ $\endgroup$ – user477343 Oct 27 '17 at 5:38
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    $\begingroup$ The expression for n = 2 should read: $(6t^3 + 1)^3 + (- 6t^3 + 1)^3 + (- 6t^2)^3 = 2$ That is, the 2nd sign in the 2nd pair of parentheses should be + rather than -. $\endgroup$ – Chris Manning Nov 26 '17 at 5:37
  • $\begingroup$ @joro: Using Huisman's results, one finds that $$x^3+y^3+z^3 = 972$$ already has $96$ solutions in the "small" finite range that he searched. Similarly for other $n$. It is tempting to speculate that, if the range were infinite, then it and others in fact have infinitely many solutions like $n=1$. $\endgroup$ – Tito Piezas III Dec 30 '17 at 3:17
  • $\begingroup$ @TitoPiezasIII What if in general the solutions are exponentially growing? $\endgroup$ – joro Dec 30 '17 at 9:15
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For $n\equiv \pm 4\pmod{9}$ there is no solution to $(1)$. Otherwise, for $n\ge 1$, it is conjectured that there are always solutions, even infinitely many. There are no analytic results, but heuristics suggest that given $n$, not $0$ or $\pm 4\pmod{9}$, solutions should occur infinitely often, asymptotically $c\log(N)$ solutions in $|x|,|y|,|z|<N$, see papers of Conn and Vaserstein.

The topic has been discussed quite frequently, see also sum of three cubes and parametric solutions, Are nontrivial integer solutions known for $x^3+y^3+z^3=3$?, Efficient computation of integer representation as a sum of three squares, etc. For a collection on polynomial parametric solutions, see https://sites.google.com/site/tpiezas/010.

See also the Euler-Binet solutions to $x^3+y^3=z^3+w^3$, $$ x = 1 − (p − 3q)(p^2 + 3q^2), $$ $$ y = −1 + (p + 3q)(p^2 + 3q^2), $$ $$ z = (p + 3q) − (p^2 + 3q^2)^2, $$ $$ w = −(p − 3q) + (p^2 + 3q^2)^2. $$

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  • $\begingroup$ Thank you. So unless n is a cube or twice a cube no infinite solutions are known? $\endgroup$ – joro Aug 8 '13 at 8:58
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https://sites.google.com/site/tpiezas/010 cites the following polynomial identity $$(m^3-3^6n^9)^3+(-m^3+3^5mn^6+3^6n^9)^3+ (3^3m^2n^3+3^5mn^6)^3=m(3^2m^2n^2+3^4mn^5+3^6n^8)^3$$ valid for any $n$ and $m$.

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  • $\begingroup$ Thanks. But the RHS varies as m,n vary so this doesn't represent fixed n? $\endgroup$ – joro Aug 8 '13 at 9:56
  • $\begingroup$ Yes, you are right. This identity does not produce infinitely many integer solutions. It only proves Ryley's Theorem that "any non-zero rational number $m$ is the sum of three rational cubes in an infinite number of non-trivial ways.” $\endgroup$ – Zurab Silagadze Aug 8 '13 at 10:08
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As stated by Dietrich Burde, it is known that there is no solution for $n\equiv \pm 4 \pmod{9}$, and conjectured that there are infinitely many solutions otherwise.

A cryptic aspect is that it is not even known that there exists one solution for all $n \not\equiv \pm 4 \pmod{9}$.
Today the smallest number for which the problem is open is $n=33$.

Here is a (non-exhaustive) history of the latest solutions found for $n \le 100$ (see here and there):

(1960s)

  • $87 = 4271^3 – 4126^3 – 1972^3$
  • $96 = -15250^3 + 13139^3 + 10853^3$
  • $91 = 83538^3 – 67134^3 – 65453^3$
  • $80 = -112969^³ + 103532^³ + 69241^³$

(1990s)

  • $39 = -159380^³ + 134476^³ + 117367^³$
  • $75 = – 435203231^³ + 435203083^³ + 4381159^³$
  • $84 = 41639611^³ – 41531726^³ – 8241191^³$

(2000s)

  • $30 = 2220422932^3 – 2218888517^3 – 283059965^3$
  • $52 = -61922712865^³ + 23961292454^³ + 60702901317^³$
  • $74 = −284650292555885^3 + 66229832190556^3 + 283450105697727^3$

Remark: for $n \le 1000$, the problem is still open only for $33$, $42$, $114$, $165$, $390$, $579$, $627$, $633$, $732$, $795$, $906$, $921$, and $975$ (see this paper and this paper).

I've discovered this problem in this recent video of Numberphile: The Uncracked Problem with 33

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