Timeline for Chebyshev polynomials of the first kind and primality testing

Current License: CC BY-SA 3.0

11 events

when toggle format	what		by	license	comment
Nov 22, 2017 at 2:45	history	edited	მამუკა ჯიბლაძე	CC BY-SA 3.0	2 is prime
Nov 20, 2017 at 6:19	comment	added	Gerhard Paseman		I probably did mean the opposite. Yes, if n is composite, then both congruences should fail. Gerhard "Doesn't Work With Ideals Ideally" Paseman, 2017.11.19.
Nov 20, 2017 at 6:03	comment	added	მამუკა ჯიბლაძე		But you probably wanted to ask the opposite, right? Because if the mod n congruence holds, it will also hold mod x^r-1. The nontrivial fact to prove is that if mod n fails then it will also fail mod x^r-1.
Nov 20, 2017 at 5:49	comment	added	მამუკა ჯიბლაძე		@GerhardPaseman Experimentally, the mod n congruence holds iff n is prime. I believe it is not difficult to show using explicit expressions for coefficients of $T_n$,$$a_{n-2k}=(-1)^k2^{n-2k}\frac n2\frac{(n-k-1)!}{k!(n-2k)!},$$$k=0,1,...$ (all others zero).
Nov 20, 2017 at 5:41	comment	added	Gerhard Paseman		It is clear that n is prime implies the desired mod n congruence. What results do you get for composite n? In particular, are there any composite n for which the mod n congruence hold and the x^r-1 congruence fails? Gerhard "Bidirectionals Are Not Always Symmetric" Paseman, 2017.11.19.
Nov 20, 2017 at 5:23	history	edited	მამუკა ჯიბლაძე	CC BY-SA 3.0	deleted 112 characters in body
Nov 20, 2017 at 5:07	history	edited	მამუკა ჯიბლაძე	CC BY-SA 3.0	updated with results of additional calculations
Nov 19, 2017 at 19:27	history	edited	მამუკა ჯიბლაძე	CC BY-SA 3.0	added 8 characters in body
Nov 19, 2017 at 18:59	history	edited	მამუკა ჯიბლაძე	CC BY-SA 3.0	added 77 characters in body
S Nov 19, 2017 at 18:32	history	answered	მამუკა ჯიბლაძე	CC BY-SA 3.0
S Nov 19, 2017 at 18:32	history	made wiki			Post Made Community Wiki by მამუკა ჯიბლაძე