[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.

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