[seqfan] Re: Digital question

[Juan Arias de Reyna]

Here is the proof of  

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



Assume that (d does not divide b^2-1)   we want to prove that there exists n with 
d | n but not d | n_1.

(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.


(b) Let  e the least natural number with  d | b^e-1.  We will have e > 2, since 
d does not divide b^2 -1 and hence also does not divide b-1.

Let x_0=1, x_1, x_2,  x_{e-1} , defined by x_j is the rest of b^j mod d.
The usual criterion of divisibility gives us  d | n is equivalent to 
d | A_0+A_1*x_1+A_2*x_2+ ... + A_{e-1}*x_{e-1} where if  A_j = a_j+a_{j+e}+a_{j+2e}+...
are the sum of the digits of the representation of n in base b.

Given elements (z_j)_{j=0}^{e-1} in Z/(dZ) it is easy to construct a number n such that  
A_j(n)\equiv z_j mod d  and  A_j(n_1) equiv z_{e-j-1} mod d.  Therefore we only need to construct
elements z_j in Z/(dZ) such that  z_0+z_1*x_1+z_2*x_2+ ... + z_{e-1}*x_{e-1}=0 and
z_{e-1}+z_{e-2}*x_1+z_{e-3}*x_2+ ... + z_{0}*x_{e-1} != 0



Assume that e is odd. e = 2f+1  then take  z_0 = 0,  z_f = -b^f, z_{e-1}=1 and all other 
z_j = 0. Then 

z_0+z_1*x_1+z_2*x_2+ ... + z_{e-1}*x_{e-1} = -b^f x_f+x_{2f} = -b^f b^f + b^(2f) = 0
and 
z_{e-1}+z_{e-2}*x_1+z_{e-3}*x_2+ ... + z_{0}*x_{e-1} =1 - b^f x_f = 1-b^(2f) != 0
since b^e = b^(2f+1) is the first power of b  equal to 1. 

Assume now that e is even  e = 2f. Since e >2  we have  e >= 4. 
In this case take  z_0 = -b^f, z_{f-1} = b, z_f = 0,  z_{2f-1}=0. Then we will have
z_0+z_1*x_1+z_2*x_2+ ... + z_{e-1}*x_{e-1} = -b^f + b x_{f-1} = -b^f + b^f = 0
and 
z_{e-1}+z_{e-2}*x_1+z_{e-3}*x_2+ ... + z_{0}*x_{e-1} = b x_f - b^f x_{e-1} = 
b^(f+1) - b^f b^(2f-1) = b^(f+1) - b^(f-1) = b^(f-1) (b^2-1) != 0.

This ends the proof. 

Regards, 
Juan Arias de Reyna


El 23/02/2011, a las 01:05, David Wilson escribió:

> 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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