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January 28th, 2012, 07:34 PM   #1
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((my) mod n ) congruent to n-1

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 (n-1)
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January 28th, 2012, 09:02 PM   #2
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Re: ((my) mod n ) congruent to n-1

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?
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January 28th, 2012, 10:07 PM   #3
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Re: ((my) mod n ) congruent to n-1

I think what you're saying is, given m and you want to find y such that Is that right?
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January 28th, 2012, 10:19 PM   #4
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Re: ((my) mod n ) congruent to n-1

Quote:
Originally Posted by CRGreathouse
I think what you're saying is, given m and you want to find y such that Is that right?
it's and
I found that modular inverse would yield the answer . But it gives only the smallest modular number. To get the number i require it has long way to go from smallest number.
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January 28th, 2012, 10:21 PM   #5
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Re: ((my) mod n ) congruent to n-1

The modular inverse gives all the answers. To get from one to the next you add the modulus.
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