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hackandthink

A nice article with a nice equation:

42 = -80538738812075974^3 + 80435758145817515^3 + 12602123297335631^3

Douglas Adams would rejoice. (checked it in bc)

Some Background from Wikipedia:

https://en.wikipedia.org/wiki/Diophantine_set#Matiyasevich's...

"Hilbert's tenth problem asks for a general algorithm deciding the solvability of Diophantine equations. The conjunction of Matiyasevich's result with the fact that most recursively enumerable languages are not decidable implies that a solution to Hilbert's tenth problem is impossible."

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If anyone's interested in more context on that particular equation, here's an article I wrote for The Aperiodical at the time: https://aperiodical.com/2019/09/42-is-the-answer-to-the-ques...

The undecidability property proven here doesn't imply that there exists at least one Diophantine equation for which we'll never know if it's solvable or not, does it?

My understanding:

There's no algorithm to decide. But for any equation we can be lucky to find a solution or a proof that there's no solution.

But this doesn't prove that there is an equation for which we'll never know if it's solvable or not.

From my understanding, while that's is technically true, given a consistent axiomatic system (like ZFC[1], the foundation of mathematics we use) there exists a diophantine equation that can't be proven to have no solutions in that system (even though it has no solutions). This mathoverflow answer[2] gives the equation and a link to the paper that shows how to calculate the constants (the numbers are huge!).

What that means in practice is that although what you wrote is true, for some diophantine equations we'd have to come up with new axioms to be able to write a proof of the inexistence of its solutions. But then, how can we be sure that the the new axioms are consistent?

[1] I'm assuming ZFC is consistent; if it's not then it can prove anything, including the existence of solutions for any equations at all

[2] https://mathoverflow.net/a/81986

Thanks.

I'm somewhat lost, but it seems to work Gödel like.

The statement is true (equation has no solution) but we can't prove it.

See https://news.ycombinator.com/item?id=34384838, I think it disproves your last sentence — at least when assuming that all solutions and all proofs of non-existence of a solution are expressible in a shared formal language.

In [0], Carl and Moroz give an explicit polynomial in 3639528+1 variables such that: a well-formed formula is a theorem in the first order predicate calculus if and only if the polynomial parametrized by the Diophantine coding of the formula (a single natural number) has a solution in N^{3639528}.

From this, they get an explicit Diophantine equation such that: the Godel-Bernays set theory is consistent if and only if that Diophantine equation has no solutions (and thus the same is true for ZFC, since NBG is a conservative extension of ZFC).

[0] https://link.springer.com/article/10.1007/s10958-014-1830-2

Does there exist a set of yes/no problems such that:

- there's no general algorithm that can solve an arbitrary problem from the set (the whole thing is undecidable)

- each problem in isolation _can_ be solved. there's no single problem that's impossible to solve

I don’t think so, at least if you assume that each concrete solution can be expressed in finite length in a formal language with a finite alphabet, and can be mechanically checked (which is generally the case for mathematical proofs). Because then you could just enumerate all strings of that language until you find one that describes the solution to the given problem, which by your second item would be guaranteed to exist, and thus the procedure be guaranteed to terminate, contradicting your first item.

Thanks for the answer, it helps piece together the puzzle. I believe there's a problem with this reasoning:

> each concrete solution can be expressed in finite length in a formal language with a finite alphabet

Suppose that's the case, the problem is that the resulting language of all the finite proofs can still be infinite and we again cannot enumerate and check all the solutions (since the final set is infinite). Therefore we're short of a method to decide yes/no for each question.

This appears to be exactly the case in the example provided by @ykonstant

> Consider the sequence of yes/no problems P_K = {Is there a solution to Q=0 in [-K,K]^n?} parametrized by a positive integer K.

For each yes/no question, the workload is finite, but for the union of all yes/no questions, the workload is not finite.

> Suppose that's the case, the problem is that the resulting language of all the finite proofs can still be infinite and we again cannot enumerate and check all the solutions (since the final set is infinite).

The set of all strings in the language is only countably infinite (= same size as tne natural numbers), which means you will reach the solution string after a finite time (just count 1, 2, 3, ... until you reach the corresponding natural number).

What I'm saying is that if each individual problem has a solution in the form of an individual finite proof (the premise of your second point), then the above gives you a general algorithm for finding the proof for any given of those individual problems (contradicting the premise of your first point).

What this doesn't give you is a way to prove that a proof exists for all individual prohlems, because that indeed would take infinite time.

Ykonstant is correct in that your use of "undecidable" was confused; I just ignored that.

You are right, I was wrong. If there's a way to evaluate each single yes/no question in a finite number of steps, then the set of all problems is decidable, since any question can be resolved in a finite number of steps.

That would seem to imply that each problem (seen as a set of sub-problems) contains some undecidable sub-problems.. so is there something like the most primitive undecidable problem..

The way you are phrasing your question is confusing; without the parenthesis, your two statements are identical. The only meaningful way to interpret "The whole thing is undecidable" is whether you can or cannot decide all (infinitely many) statements at once.

If that is what you mean, then yes: take any undecidable Diophantine equation Q=0 of, say, n variables. Consider the sequence of yes/no problems P_K = {Is there a solution to Q=0 in [-K,K]^n?} parametrized by a positive integer K. Each of those problems is decidable in isolation, but the totality cannot be, since that would decide if Q=0 has a solution or not.

Thank you for suggesting this example.

For the undecidable Diophantine equation Q=0 of n variables - I'm assuming that the fact that someone could hit a solution randomly (by randomly generating the n variables and getting lucky) does not contradict with its undecidability.

Still I'm not clear what happens if such a solution would be randomly hit for an equation that's mapped to another hard question such as ZFC consistency. It would imply that a question such as ZFC consistency could be solved by randomly generating an answer, which seemingly doesn't make sense?

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