Maybe it's silly, but from someone who is ignorant on this topics, what are the consequences of this kind of "discoveries"?

Is it something "revolutionary" or just another small brick that will pile up until something really "revolutionary" will happen?

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arm32 12 hours ago | parent | next [-]

To me, this shows that extremely talented and qualified mathematicians (can) use frontier-level LLMs to automate their personal grind-y workloads that would otherwise (probably) take more time to accomplish with natural intelligence.

By itself, no consequence. But over time, provided we keep pumping out talented and qualified mathematicians and keep subsidizing costs, we could maybe hit a breakthrough... somewhere... that has real impact.

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sdwr 12 hours ago | parent | prev | next [-]

It's an indicator of AI progress. The solutions aren't especially revolutionary, but no person had been able to solve them after decades of collective attempts.

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hyperhello 12 hours ago | parent [-]

To be fair I don’t think there were too many people really trying to. Symbolically, one could make a parameterization of the Jacobian determinant and then brute force a solution, if one had known such a polynomial existed in only three dimensions.

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traes 9 hours ago | parent | next [-]

This is not true at all. The parameter space is absolutely MASSIVE. The counterexample is a degree 7 polynomial in 3 variables, which means 360 coefficients. There's no particular way to bound these coefficients or even the degree or number of variables apriori, but assume you somehow did. Also assume you were confident that it would work with integer coefficients bounded from -12 to 12. Now you have to iterate over 360 degrees of freedom, verify that the Jacobian is a nonzero constant, and somehow show uninvertibility of the transformation, which is not a particularly simple task.

If you searched for coefficients from -12 to 12, this would be 25^360 = 2 * 10^503 different possibilities. A common reference point is that there are 10^80 atoms in the observable universe. Sure you could probably reduce this a bit with clever tricks, but the starting point makes the method completely unviable, even with the knowledge: A) a counterexample exists, B) it's in 3 variables, C) it's in degree 7 or less, D) it's in integer coefficients, E) those coefficients are 12 or lower.

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pfdietz 7 hours ago | parent [-]

Here the search wouldn't have been chosing the coefficients independently. Note that one intermediate variable is a polynomial in the input variables, and it is used in other polynomials. A search over expressions like the ones in the counterexample would have a much smaller search space.

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traes 7 hours ago | parent [-]

1) How do you know this structure is the correct one a priori

2) You are starting at 10^500 possibilities. "Much" smaller is not enough, the order of magnitude of the order of magnitude needs to be changed.

3) You still need all of the other assumptions, which were completely unfounded

Impossible.

	▲	hyperhello 7 hours ago \| parent \| next [-]
		No, I’m not saying you would know in advance that it was possible, but sometimes you visit the crystal cave and the diamond is just sitting there, so why not work out the odds? I learned from poking around that checking the invertibility of a system in C is a much, much harder problem than I thought. Nonetheless if that were no object, let’s say coefficients from -16 to 15 (5 bits) times eight terms times choosing up to cubes (64) times three equations is searchable, especially since you have only the final combination of coefficients in the determinant. It’s not impossible to generate the equations like this Fizzbuzz style. Edit: no. 2048 possible monomials, to the 24th power, not times 24. Fine, can’t brute force it.
	▲	pfdietz 6 hours ago \| parent \| prev [-]
		If I pull out the two terms 1+xy and 3 + 4xy as new variables, then make three polynomials that are <= 3 terms in each with coefficients in the range -3 to 3, then there are something like 10^19 possibilities. Multiply this by the various simple possibilities for the definitions of those two new variables. The coefficients are mostly 1, so biasing toward that would make it much faster.

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Legend2440 12 hours ago | parent | prev [-]

Oh yes there were. The Jacobian conjecture is "notorious for the large number of published and unpublished false proofs which turned out to contain subtle errors."

It's not quite the Reimann hypothesis, but many prominent mathematicians have spent years working on this problem. Yitang Zhang wrote his PhD thesis on it.

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hyperhello 9 hours ago | parent [-]

I shouldn’t, but:

F1 = x^3y^3z + 3x^2y^4 + 3x^2y^2z + 7xy^3 + 3xyz + 4y^2 + z

F2 = 3x^3y^2z + 9x^2y^3 + 6x^2yz + 12xy^2 + 3xz + y

F3 = -x^3z - 3x^2y + 2x

That’s the counterexample. Low integer coefficients, power 7 in three variables. If someone said it was there, couldn’t we all have written a pretty simple brute force solution for the search space, especially with the constraints that the symbolic determinant had to cancel to a constant?

	▲	traes 9 hours ago \| parent \| next [-]
		Honestly just try it. You'll figure out the problem very quickly.
	▲	elisbce 8 hours ago \| parent \| prev [-]
		I don't think you even understand the problem. The determinant needs to be a non-zero constant AND you need to prove that particular map is not globally injective, meaning you have to find at least two points mapping to the same value. Of course it looks easy when someone shows you the counterexample.

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fragmede 12 hours ago | parent | prev | next [-]

Practically, from this specific one? Nothing, it's very much a math thing. It's like art or music at this level. Are there consequences to a van Gogh?

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hyperhello 12 hours ago | parent | prev | next [-]

[dead]

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rickypp 13 hours ago | parent | prev [-]

"Hey Fable, please generate me the next 1000 undiscovered bitcoin hashes"

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echelon 12 hours ago | parent [-]

You're joking, but perhaps LLMs will find a way to mathematically break the complexity of factorization.

Maybe they'll find a solution where P=NP.

That could really throw a wrench into the whole internet thing.

It seems they need an expert human driver for now.

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ConceptJunkie 12 hours ago | parent [-]

I'm sorry, I can't do that, but here is the design for a stable quantum computing platform that should allow you to generate those keys yourself...

	▲	NitpickLawyer 12 hours ago \| parent [-]
		Some materials are readily available on eMazon and aBay, so I've taken the liberty of ordering those for you. Your credit card bill will be a bit high this month, but it'll be worth it. There weren't any sellers for the advanced EUV lithography machines, so I've hacked into the only place on earth that makes them, changed their records and had them ship it to you. Expect to receive a "pinball machine" from Amsterdam, soon. I've instructed the roomba connected to the local network to start assembling stuff while we wait for the other materials. Oh, and you're gonna need a new toaster.