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Old 2018-01-22, 15:26   #12
lukerichards
 
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"Luke Richards"
Jan 2018
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Quote:
Originally Posted by science_man_88 View Post
2)see formula from 1, M_m= 2^{m-n}M_n+M_{m-n} q divides the first term, and so q must divide into the second term in order to divide the LHS.
This is very helpful, thanks! It took me a while to get my head around (best described as I was looking at it from the wrong direction - starting with M_m and trying to get to M_n through division and ending up with some horrible negative powers of 2.

So I've got: (avoiding using the spoiler tags, to keep the latex operational, so spoiler below...)








m - n = x

M_m = M_{n + x}

M_{n + x} = {2^x} \cdot M_n  + M_x

Given M_{n + x} mod q = {2^x} \cdot M_n mod q = 0

This simplifies to:

0 mod q = 0mod q + M_{m - n}

Therefore M_{m - n} = 0 mod q

Last fiddled with by lukerichards on 2018-01-22 at 15:37 Reason: consistency in use of LaTex
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Old 2018-01-22, 16:45   #13
Nick
 
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Dec 2012
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Quote:
Originally Posted by lukerichards View Post
So I've got:...
Yes, that's great.
Now the challenge is to use what you have so far to do the 3rd part!
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