[seqfan] Re: Digital question

David Wilson davidwwilson at comcast.net
Wed Feb 23 01:05:40 CET 2011


Very good. Now show

(d does not divide b^2-1) => there exists n with d | n but not d | n_1.


----- Original Message ----- 
From: "Vladimir Shevelev" <shevelev at bgu.ac.il>
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Sent: Tuesday, February 22, 2011 7:33 AM
Subject: [seqfan] Re: Digital question


> From b^(t-1)*n-n_1==0 (mod d) we conclude that d | n => d | n_1. Since 
> {n_1 is inv. n}<=>
> {n is inv. n_1}, then also b^(t-1)*n_1-n==0 (mod d) and we conclude that d 
> | n_1 => d | n.
>
> Regards,
> Vladimir



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