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I was reading this answer, which says that:

In his Master's Thesis, Merlin Carl has computed a polynomial that is solvable in the integers iff ZFC is inconsistent. A joint paper with his advisor Boris Moroz on this subject can be found at http://www.math.uni-bonn.de/people/carl/preprint.pdf.

Note that the link is dead. Emil Jeřábek provided an alternate link here: A polynomial encoding provability in pure mathematics (outline of an explicit construction).

That phrase "solvable in the integers iff $\mathsf{ZFC}$ is inconsistent". If $\mathsf{ZFC}$ is inconsistent, then of course the polynomial is solvable in the integers - every statement in the model is true! So it seems to be just a fancy way of saying that the polynomial is not solvable in the integers.

I believe I'm missing some subtleties here, so I would like to have someone address this confusion of mine.

Clement Yung
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  • Maybe ZFC is inconsistent but a weaker theory in which you can still do most math (i.e. Zermelo set theory ZC) is consistent? – Monroe Eskew Mar 22 '22 at 09:48
  • Good link: https://projecteuclid.org/journals/bulletin-of-the-belgian-mathematical-society-simon-stevin/volume-20/issue-1/A-polynomial-encoding-provability-in-pure-mathematics-outline-of-an/10.36045/bbms/1366306724.full – Sean Eberhard Mar 22 '22 at 09:55
  • It means that there is an explicit polynomial $P$ "encoding provability of inconsistency" in the sense that in principle you can check whether ZFC is consistent by just checking whether $P(z) = 0$ for all integers $z$ in turn. This is completely different from just saying "hey check it out $x^2 + 1 = 0$ doesn't have solutions in integers, assuming ZFC is consistent of course, because if not then anything goes". – Sean Eberhard Mar 22 '22 at 10:03
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    You are mixing up theory and metatheory. It is perfectly possible to be in a model of ZFC in which the statement “ZFC is inconsistent” is true. (Moreover, I haven’t read the thesis, but the equivalence is probably proved in something much weaker than ZFC anyway, like PA.) The statement means what it literaly says: there is a polynomial $p\in\mathbb Z[\vec x]$ such that $(\exists\vec x\in\mathbb Z,p(\vec x))\leftrightarrow\neg\mathrm{Con_{ZFC}}$. – Emil Jeřábek Mar 22 '22 at 10:04
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    There is an explicitly-known Turing machine that halts iff ZFC is inconsistent. This is a problem of a similar nature. Imagine a Turing machine that will check values of the polynomial, and will halt when it finds a root... – David Roberts Mar 22 '22 at 10:09
  • @DavidRoberts Note that the link you inserted is one of the two that I, apparently, posted in the comments back then. It’s the one that “only shows the first page of the thesis” as the OP writes, except that the full PDF is, in fact, available there for download. The OP must have missed that. – Emil Jeřábek Mar 22 '22 at 10:32
  • @EmilJeřábek sorry! I didn't actually go and check the old post (that should be updated, BTW, if it hasn't already), I just went and grabbed the doi link from the Project Euclid page linked in Sean's comment. – David Roberts Mar 22 '22 at 11:06
  • @Clement I appreciate you are cleaning up your question, but you did remove a useful piece of information I included for the sake of readers, namely the actual title of the paper in question! – David Roberts Mar 22 '22 at 11:07
  • @DavidRoberts I agree including the title is better. I've edited my question. – Clement Yung Mar 22 '22 at 11:09
  • The definition of "solvable in the integers" does not invoke ZFC. – Steven Landsburg Mar 22 '22 at 13:56
  • Since I see that the dead link was discussed above, I'll just add mention that a snapshot from 2013 in Wayback Machine shows at that URL a preprint called
    On a Diophantine representation of the predicate of provability. That seems to be the paper published with DOI 10.1007/s10958-014-1830-2.     – Martin Sleziak Mar 29 '22 at 01:40

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Edit: After I wrote this, the paper was posted in the commentary. What I wrote below is what is meant, however they use $\mathsf{GBC}$ instead of $\mathsf{ZFC}$. Since these two theories are equiconsistent and have the same first order consequences, this technical difference is immaterial.

However, when people say things like this there are a number of ways to interpret it, none of which are trivial. If I had to guess, probably what is meant is something like ``there is a polynomial $p(x)$ (often concretely computed) and $\mathsf{ZFC}$ proves ($\exists x \in \mathbb Z$ $p(x) = 0$ iff $\mathsf{ZFC}$ is inconsistent) ". By the second incompleteness theorem even if $\mathsf{ZFC}$ is consistent it cannot prove this fact so the equivalence is not trivial. Moreover, there are models on $\mathsf{ZFC}$ which think $\mathsf{ZFC}$ is consistent, and in such models there are no integer solutions to $p(x)$ and other models of $\mathsf{ZFC}$ which think that $\mathsf{ZFC}$ is inconsistent and in such models some (necessarily nonstandard) integer will satisfy $p(x)$.

The point is usually that the kinds of coding that lead to the incompleteness theorems can in fact be coded into surprisingly concrete polynomials. In the case of $\mathsf{PA}$ there are many examples of such coming from MRDP/Matiyasevich's theorem. In the wikipedia article on diophantine equations this is mentioned: https://en.wikipedia.org/wiki/Diophantine_set, see ``further applications".

Corey Bacal Switzer
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    I think it's worth noting that the theory that proves that the polynomial has a root iff $\mathsf{ZFC}$ is inconsistent can actually be much weaker than $\mathsf{ZFC}$ or even $\mathsf{PA}$. – James Hanson Mar 22 '22 at 19:19