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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.
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]
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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:
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].
This conjecture is recorded as Problem 3.64(a) in the Kirby problem List [2].
The conjecture remains unresolved despite of decades of research, for instance:
The 2015 paper by Michel Boileau, Steven Boyer, Radu Cebanu, Genevieve S. Walsh (cf. Conjecture 1.1). [3]
The 2020 paper by Eric Chesebro, Jason DeBlois, Neil R Hoffman, Christian Millichap, Priyadip Mondal, William Worden (cf. Conjecture 1.1). [4]
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]