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i want to add an unsolved math question which is (12 45 ∏ 61 ​

35)! 2601:603:4C7F:B6D0:68E7:C8AD:7A34:6A7 (talk) 17:07, 17 June 2024 (UTC)[reply]

What does it mean? —Tamfang (talk) 23:49, 17 June 2024 (UTC)[reply]

 Not done for now: Critical lack of explanation why this should be included in the list, and no sources. ABG (Talk/Report any mistakes here) 23:55, 17 June 2024 (UTC)[reply]

Like the Erdős-Heilbronn conjecture. 2405:201:5502:C989:D1F5:2160:CCE8:4F0A (talk) 05:16, 15 July 2024 (UTC)[reply]

 Done Any other ones you noticed? GalacticShoe (talk) 06:34, 15 July 2024 (UTC)[reply]

1. Carbrickscity conjecture

The conjecture asks, whether Graham's number - 4 is a prime.

Graham's number: a power of 3

Graham's number - 1: even

Graham's number - 2: a multiple of 5

Graham's number - 3: an even multiple of 3

Graham's number - 4: unknown

2. repunit power conjecture

There are infinitely many cubes of the form 3 mod 4.: 27, 343, 1331, 3375, 6859, 12167, 19683, 29791, 42875, 59319, 79507, 103823, 132651, 166375, 205379, ... (A016839)

There are infinitely many fifth powers of the form 3 mod 4.: 243, 16807, 161051, 759375, 2476099, ... (A016841)

This goes on with any odd exponent.

So, the conjecture asks, whether a repunit other than 1 can be equal to aⁿ, where a is an integer and n is odd and greater than 1.

It is sure, that a repunit other than 1 can never be a square, because squares can never be of the form 3 mod 4, while repunits other than 1 are always of the form 3 mod 4. 94.31.89.138 (talk) 19:53, 28 July 2024 (UTC)[reply]

This is not the place to pose new conjectures. All content here, as in all Wikipedia articles, must be based on reliably-published sources. If you have citations for sources for conjectures to be added, they can be listed here. If not, then they need to be published elsewhere before they can be considered here. —David Eppstein (talk) 20:28, 28 July 2024 (UTC)[reply]

Add the open problems from here, including whether PP implies AC, whether WPP implies AC, and whether the Schröder–Bernstein theorem for surjections implies AC. These are some of the oldest open problems in set theory. 50.221.225.231 (talk) 16:02, 26 November 2024 (UTC)[reply]

Request:

I want to request for “Neumann-Reid Conjecture” to be added to the list of unsolved problems in “Topology”.

The statement of the conjecture:

“The only hyperbolic knots in the 3-sphere which admit hidden symmetries are the figure-eight knot and the two dodecahedral knots.”

Background and resources:

1. In 1992 Neumann and Reid established that hyperbolic knot complements have hidden symmetries if and only if they cover rigid-cusped orbifolds, in the same paper they further questioned whether any examples exist beyond the three known: the complements of the figure-eight knot and the two dodecahedral knots (cf. Section 9, Question 1)[1].
2. This conjecture is recorded as Problem 3.64(a) in the Kirby problem List [2].
3. The conjecture remains unresolved despite of decades of research, for instance:
   1. The 2015 paper by Michel Boileau, Steven Boyer, Radu Cebanu, Genevieve S. Walsh (cf. Conjecture 1.1). [3]
   2. The 2020 paper by Eric Chesebro, Jason DeBlois, Neil R Hoffman, Christian Millichap, Priyadip Mondal, William Worden (cf. Conjecture 1.1). [4]

ArshiGh (talk) 01:29, 3 February 2025 (UTC)[reply]

 Not done: it's not clear what changes you want to be made. Please mention the specific changes in a "change X to Y" format and provide a reliable source if appropriate. Your request is valid and well written. However, it may not be added because of the xy policy.(3OpenEyes' communication receptacle) | (PS: Have a good day) (acer was here) 09:40, 4 February 2025 (UTC)[reply]

Remove the Carathéodory conjecture from this list as it has been solved in 2024 by Guilfoyle and Klingenberg (full citations contained in Carathéodory conjecture). Boundary Condition (talk) 18:20, 23 February 2025 (UTC)[reply]

I'm not an expert on this topic, but reading the article it seems that the solution was for a specific case, not a fully general solution. Can you elaborate a bit? PianoDan (talk) 00:27, 25 February 2025 (UTC)[reply]

So our article on the Carathéodory conjecture has the following formulation:

"The conjecture claims that any convex, closed and sufficiently smooth surface in three dimensional Euclidean space needs to admit at least two umbilic points."

The way I interpreted the result is that they apparently showed this to be true when "sufficiently smooth" is . Based on our section on Hölder spaces it appears this means that the surface is and its third-order partial derivatives all satisfy the Hölder condition of order .

This would confirm the original conjecture in its literal formulation of "sufficiently smooth". Since the Hölder condition is weaker than continuous differentiability, surfaces all work, but some surfaces may not. There still does seem to be the question of if it's possible to lower the smoothness class required to the lowest possible though. I will also need someone else to confirm whether my interpretation is correct. GalacticShoe (talk) 16:18, 25 February 2025 (UTC)[reply]

This is correct - the original Conjecture, first reported in 1924, made no mention of the exact degree of smoothness of the surface. The phrase "sufficiently smooth" has been inserted later - the minimal smoothness required for the Conjecture to make sense is . The 1940's proof for real analytic surfaces depended very heavily on real analyticity and, as such, was a special case of the Conjecture. The big jump is from real analytic to smooth, and proving it for any degree of smoothness is sufficient to confirm the original Conjecture. Boundary Condition (talk) 08:40, 26 February 2025 (UTC)[reply]