20120129, 04:36  #1 
Apr 2011
7 Posts 
((my) mod n ) congruent to n1
If given a 'n' value and m = floor ( squareroot(n) )
then is there any way to find the value of 'y' , such that ((m*y) mod n) is congruent to (n1) 
20120129, 06:01  #2 
Apr 2011
7 Posts 
with the help of a friend
i figured out that, if m is the divisor of n, it wont be possible to get a solution . But what about the other values? 
20120129, 11:30  #3  
Apr 2010
2^{2}×37 Posts 
Quote:
Last fiddled with by ccorn on 20120129 at 11:41 
