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Apr 2, 2023 at 5:56 history edited Bjørn Kjos-Hanssen CC BY-SA 4.0
added 1 character in body
Apr 2, 2023 at 5:56 comment added Bjørn Kjos-Hanssen There is also $(d p)=(2 3) $ :)
Mar 29, 2023 at 11:12 review Suggested edits
Mar 29, 2023 at 13:16
Mar 29, 2023 at 9:53 answer added user178594 timeline score: 0
Apr 12, 2014 at 19:07 comment added Włodzimierz Holsztyński @TMA, if you posted your question about $f(p)$, I would welcome it (while "similar interesting" seems vague and too encompassing).
Apr 12, 2014 at 12:20 vote accept Włodzimierz Holsztyński
Apr 12, 2014 at 11:37 answer added user46855 timeline score: 3
Apr 12, 2014 at 9:50 answer added Aravind timeline score: 9
Apr 12, 2014 at 8:17 history edited Włodzimierz Holsztyński CC BY-SA 3.0
A more complete statement (within the acknowledgement).
Apr 12, 2014 at 7:52 history edited Włodzimierz Holsztyński CC BY-SA 3.0
Upper case
Apr 12, 2014 at 5:36 history edited Włodzimierz Holsztyński CC BY-SA 3.0
better grammar, I guess.
Apr 12, 2014 at 5:26 comment added The Masked Avenger You might prefer the following. Let f(p) be the power of 2 that divides precisely (O_p)^2 - 1. Is f(p) unbounded as a function of p? Similar interesting questions around f(p) could be posed.
Apr 12, 2014 at 5:26 history edited Włodzimierz Holsztyński CC BY-SA 3.0
Motivation
Apr 12, 2014 at 5:19 comment added The Masked Avenger Likely yes. For large p Fermat's little theorem will determine the character of the primes dividing 2^p +- 1. You can check this out using Carmichael's tables to shoa no other small solutions.
Apr 12, 2014 at 5:11 history asked Włodzimierz Holsztyński CC BY-SA 3.0