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**yourskadhir**- Replies: 1

Hi,

For any odd composite 'N', let u = (N-1)/2, v = u+1, then u^2(mod p) = v^2(mod p) if and only if 'p' is a factor of 'N'.

For more info please visit kadinumberprops.blogspot.in

**yourskadhir**- Replies: 0

In order to factorize an odd composite N = pq where q >p, p and q either prime or composite through Fermat's factorization.(1) Many of as think that the difference between the q and p is the reason behind the time complexity.(2) And moreover we have logic that any number with least difference like 100, 200, etc., will be factored easily compared 1000, 10000, etc., Both logics (1) and (2) are wrong. Reason please follow the url

kadinumberprops.blogspot.in

**yourskadhir**- Replies: 0

The logic that odd composite with least difference will be factored easily and large difference would factored hardly is wrong. B'coz whatever be the difference between the factors their exist Best Fermat Factors to make the Fermat factorization easier. Please follow the following url to know more.

kadinumberprops.blogspot.in

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