[seqfan] Re: Digital question
Juan Arias de Reyna
arias at us.es
Thu Feb 24 10:21:17 CET 2011
My reasoning leaves the case in which d | b^k for some k ( the case in which my previous d_1 = 1)
In this case if k is the least number choose n = b^k then n_1 = 1 and there is no problem.
Best regards
Juan Arias de Reyna
El 24/02/2011, a las 09:58, Juan Arias de Reyna escribió:
> (a) We may assume that gcd(d,b)=1. In other case d = d_1*d_2 with gcd(d_1,b)=1
> d_2 | b^k for a suitable k. Now (d_1 does not divide b^2-1) and if we find
> n with d_1 | n but not d_1 | n_1. Take m=n*b^k it is clear that d | m and
> m_1 = n_1, so that d does not divide m_1.
More information about the SeqFan
mailing list