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The Diophantine equation $$x^3 + y^3 + z^3 = 42$$ was recently solved by Booker and Sutherland: Sum of three cubes for 42 finally solved.

Is there a clean partition of the form of those polynomial equations all of which do have integer solutions, and those that are known to be undecidable (following the negative solution of Hilbert's 10th)? Or—in the absence of a clean partition—can at least the equations be partitioned into: $$\{ \textrm{solvable, undecidable, unknown} \}$$

Am I correct that the status of $x^3 + y^3 + z^3 = c$ is unknown except for certain values of $c$? I ask this naively; not my expertise.

Related MO question: What are the solutions of this Diophantine equation?.

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The negative solution of Hilbert's tenth problem means that there is no algorithm that would recognize the solvable polynomial equations. Hence the "clean partition" you are looking for does not exist.

Of course, one can partition the polynomials any way one likes. The partition solvable/non-solvable/undecidable is a legitimate one, except that one "does not know" which polynomials belong to the individual parts.

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  • $\begingroup$ Thank you. I thought e.g., linear Diophantine equations and Pell's equation fell into the "solvable" class, or at least the "understood" class. $\endgroup$
    – Joseph O'Rourke
    Commented Feb 15, 2020 at 14:02
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    $\begingroup$ @JosephO'Rourke: Yes, there are well-understood classes of Diophantine equations. However, Matiyasevich's theorem states, in a sense, that we shall never fully understand which equations are solvable/non-solvable/undecidable. Just as we shall never fully understand which statements in number theory are provable/falsifiable/undecidable. We make progress from day to day (this is our job), but we are walking on an infinite path. $\endgroup$
    – GH from MO
    Commented Feb 15, 2020 at 16:14
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    $\begingroup$ "we are walking on an infinite path"---Well phrased! $\endgroup$
    – Joseph O'Rourke
    Commented Feb 15, 2020 at 16:21
  • $\begingroup$ @MarkSapir: Matiyasevich's theorem also implies that there exists polynomials for which solvability is undecidable within ZFC (assuming ZFC is consistent). Indeed, assume that for each polynomial solvability is decidable within ZFC. Then, given an arbitrary polynomial, you can go through the first order statements in ZFC one by one, and after a while you will arrive at a proof or disproof of solvability (because a proof in ZFC is a first order statement in ZFC). This is a general algorithm for solvability, but Matiyasevich proved that such an algorithm does not exist. Contradiction. $\endgroup$
    – GH from MO
    Commented Feb 15, 2020 at 16:27
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    $\begingroup$ @AndrésE.Caicedo: It is conjectured that every integer not congruent to $\pm 4$ modulo $9$ is a sum of three integral cubes. There are a few positive results, but otherwise the status of these equations is unknown (including their possible independence of ZFC). $\endgroup$
    – GH from MO
    Commented Mar 14, 2020 at 22:43

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