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A famous corollary of Matiyasevich's theorem is that there exists a Diophantine equation such that it is undecidable (under some recursively axiomatizable theory $T$, such as ZFC) whether that equation has any natural-number solutions. I will somewhat sloppily describe such a Diophantine equation as "undecidable" (with respect to $T$) for brevity.

I've never actually seen an explicit example of an undecidable Diophantine equation. The closest that I've seen is in this answer to another Math Overflow question, which provides an example that's almost explicit, but contains one unspecified parameter $K$, whose value depends on the recursively axiomizable theory $T$. However, that answer claims that $K$ can be effectively computed from the axioms of $T$.

A. Are there any simpler explicit Diophantine equations that are known to be undecidable with respect to some common axiomatic system (ideally ZFC, or else ZF or ZFC + CH or something)?

B. For, say, ZFC (or a similar theory, as described in question A), which of the following mutually exclusive statements best captures how well we can actually calculate the value of $K$?

  1. An explicit numerical value of $K$ has been reported in the literature.
  2. We know how to calculate a valid choice of $K$ and could actually do it in the real world if we really wanted to, but doing so would be too tedious for anyone to have actually bothered to do it so far.
  3. We know how to calculate a valid choice of $K$ in principle, but doing so would be impossible in practice - e.g. it would require many, many years of labor, or more computational resources than are currently available with existing supercomputers.
  4. We know how to calculate a valid choice of $K$, but it isn't clear whether doing so would be feasible in practice (so it isn't clear whether case #2 or #3 above is correct).
  5. We know that $K$ can be effectively computed from the axioms of $T$, but we don't know how to actually do so, so we cannot find a valid choice of parameter $K$ with our current mathematical knowledge.
tparker
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    For A, see https://mathoverflow.net/a/32914 – Emil Jeřábek Aug 14 '22 at 07:08
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    It would be a tedious and unrewarding task to write a program that computes such a number. But why would we do that? What is your motivation? – Andrej Bauer Aug 14 '22 at 09:22
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    The problem here begins with the word “corollary” — if you use Matiyasevich’s theorem as a black box then it doesn’t look constructive. But of course the theorem is fully constructive, and the proof (explained well in multiple places) provides the algorithm. –  Aug 14 '22 at 09:33
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    The statement is really a corollary of Matiyasevich’s theorem and Gödel’s incompleteness theorem: see this question. (That is, the “undecidable” in Matiyasevich’s theorem is one type of undecidable, and the “undecidable” in terms of ZFC is another). – Carl-Fredrik Nyberg Brodda Aug 14 '22 at 09:35
  • @MattF. So it sounds like we can eliminate option #5, but we don't know which of options #1-4 are correct? – tparker Aug 14 '22 at 18:13
  • @EmilJeřábek Where in the cited paper does the polynomial appear? It's a rather long paper with a lot of complicated polynomials in it, and it isn't clear from a quick skim which is the desired polynomial in question. – tparker Aug 14 '22 at 18:17
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    We can rule out both 1 and 5. The details of 2-4 are not clear to me given the particular family of Diophantine equations specified in the question. For undecidable Diophantine equations not restricted to that family I think we’re in case 2. –  Aug 14 '22 at 19:10
  • @AndrejBauer The MRDP theorem can create the impression that there is nothing left to be done, but as a long-term research goal, it would be interesting to know exactly which Diophantine equations we can hope to solve and which we cannot. Getting some sense of the size of the current upper bound is a first step. Compare with the work of Aaronson and Yedidia, who computed an explicit upper bound on the size of a Turing machine that ZFC cannot prove halts. – Timothy Chow Aug 15 '22 at 03:14
  • @TimothyChow: I agree it's an interesting problem, and in fact once upon a time I made an experiment with John Langford and Harvey Friedman, pushing the lower bounds. The experience from there was that one very quickly hits very difficult problems, so MRDP-style constructions are really quite far away from the lower bounds – but that's hard to prove, and that's why it's interesting. If that's the motivation, then the title of the question is misleading. – Andrej Bauer Aug 15 '22 at 09:27
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    Implementing Matiyasevich's theorem is essentially the task of writing a compiler, where the target language is not machine code, but Diophantine equations. If you want someone to do it, you should pay money for it. – Robert Furber Aug 20 '22 at 05:05
  • Since there's been work calculating what Busy Beaver values are not decidable in ZFC, it might be easier to use one of those with a small universal Turing machine to search for them, and use Matiyasevich's construction on that. – JoshuaZ Nov 27 '22 at 12:28

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Turning the proof that there exists a Diophantine equation encoding eg the consistency of $\mathrm{ZFC}$ into a program that actually computes the polynomial would be a bit tedious, but there should not be any significant obstacles. Whether running said program would actually yield an answer, some kind of overflow error, or whether we'd simply get too bored waiting for it to do anything at all is less clear to me.

What I am certain of is that if we do get an answer, it would be sufficiently monstrous that we couldn't gain any insight from it. (And I assume that people generally agreeing with me is the reason why people don't really try it out.)

Getting a "small" undecidable Diophantine equation will require much more work. At any stage of the encodings, one should look for ways to be more efficient.

To better understand the boundary between decidable and undecidable Diophantine equations, it is probably more promising to look at properties of the polynomials that allow for decision procedures vs those properties polynomials obtained from the Turing machine translations have.

Arno
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