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2018-01-22, 15:26   #12
lukerichards

"Luke Richards"
Jan 2018
Birmingham, UK

25·32 Posts

Quote:
 Originally Posted by science_man_88 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 = $0$mod 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

2018-01-22, 16:45   #13
Nick

Dec 2012
The Netherlands

3×52×19 Posts

Quote:
 Originally Posted by lukerichards 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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