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gcd = {\displaystyle {\sqrt {N}}} N a For an integer n, we want a and b such as: n = a2 - b2 = (a+b) (a-b) where (a+b) and (a-b) are the factors of the number n 2 Input: N = 105327569 d − u v ) is known, then the rational number ; if neither factor equals one, it is a proper factorization of N. Each odd number has such a representation. y u Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. > m 1 ( a close, link d Fermat's method works best when there is a factor near the square-root of N. If the approximate ratio of two factors ( . {\displaystyle d/c} = / 2 , so the number of steps is approximately N u = Going up from c ( v {\displaystyle \gcd(N,cv)=c} d = x Fermat's method gives factors quickly when they are near the root of the number. + . N Given a sequence of a-values (start, end, and step) and a modulus, one can proceed thus: But the recursion is stopped when few a-values remain; that is, when (aend-astart)/astep is small. It is apparent that only the 4 from this list can be a square. 80 {\displaystyle a-b=N/(a+b)} Otherwise it is more work than trial division. a 1 ), one needs = − N For example, if = {\displaystyle b=21} But observe that if N had a subroot factor above {\displaystyle N=\beta z+r} Experience, Take the value of sqrt(n) as a and increment the number until and unless a number b is found such that. v How is the time complexity of Sieve of Eratosthenes is n*log(log(n))? R. Lehman devised a systematic way to do this, so that Fermat's plus trial division can factor N in + brightness_4 + {\displaystyle a+b} − ) I have come up with the Python code below, which uses the gmpy2 module for accuracy. 2 The end result is the same: a difference of square mod n that, if nontrivial, can be used to factor n. https://en.wikipedia.org/w/index.php?title=Fermat%27s_factorization_method&oldid=984206978, Articles with unsourced statements from January 2015, Articles with unsourced statements from January 2020, Creative Commons Attribution-ShareAlike License, This page was last edited on 18 October 2020, at 20:39. {\displaystyle O(N)} , Fermat's method would have found it already. {\displaystyle \mathbb {Z} _{\beta }} N {\displaystyle {\left(4N\right)}^{1/4}} x N q / − / The number gets factored down … is a good estimate for the smaller of a divisor pair. u / But if N has a factor close to its square root, the method works quickly.

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