From: Jim choate <ravage@bga.com>
Date: Fri, 17 Jun 94 16:06:01 PDT
To: cypherpunks@toad.com
Subject: Some observations on xMODn...
Message-ID: <199406172305.SAA21693@zoom.bga.com>
MIME-Version: 1.0
Content-Type: text


I propose to clarify with a little mathematics as best I can what I
was, and am, asking...

To those this material appears obvious too please feel free to delete.

As I understand it MOD is a function which returns the remainder of a 
number (x) when divided by another number (n). As an example:

5mod3=2    ie 3 will go into 5 a single time and there will be a left over
              of 2.

11mod3=2   ie 3 will go into 11 a total of 3 times and there will be 2
              left over.

I propose there is a periodicity in the mod function:

n        0  1  2  3  4  5  6  7  8  9  10  11  12

nmod5    0  1  2  3  4  0  1  2  3  4  0   1   2

this can be simplified into a generic formula for a sequence:

       rem = (kn)+i  |big #          |big #
                     |               |
                     |i=0            |k=0

What this formual does is give you the sequence of any given remainder
for xmodn.

In a generic algorithm it appears as such:

     n = some number
     for k = 0 to "some really big number"
          for i = 0 to "some really big number"
               rem=(k*n)+i
          next i
     next k

From p.282 on Schneier the RSA encryption algorithm is given as:
           e
     c  = m (mod n)
      i    i

In my notation this reduces to:

     rem = (kn)+i  |          |
                   |          |
                   | n=0      |i=0

What I am asking is that since the numbers we are looking at are  very 
large there should (to the way I am thinking at the moment) some means
of detecting a sequence of patterns of periodicity related to the difference
between the actual key and the key we just select randomly. 

Specificaly what I am asking for is some reference to some work in this area.
I don't know what it is called, it doesn't appear in any books that I have 
looked at.  

Thanks for any help you may be able to provide...