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By : jeroen van gorkum
Date : November 17 2020, 11:52 AM
Hope that helps math.sqrt() uses standard IEEE 64-bit values. It can only calculate accurately for arguments less than ~2**53. Your value for n is larger than that.
If you want exact integer square roots for large numbers, I would recommend gmpy2. code :
``````import gmpy2

n = 590632926550117049

a = gmpy2.isqrt(n) + 1
b2 = a * a - n

while not gmpy2.is_square(b2):
a = a + 1
b2 = a * a - n

b = gmpy2.isqrt(b2)

p = a + b
q = a - b

print("p =", p)
print("q =", q)
`````` ## Fermat factorization method limit

By : user1840172
Date : March 29 2020, 07:55 AM
wish helps you The (< limit y) check is not necessary because, worst-case, the algorithm will eventually find this solution: ## Speeding up my Fermat Factorization function (Python)

By : Nasusu
Date : March 29 2020, 07:55 AM
Hope that helps Consider rewriting this script to use only integers instead of arbitrary precision floats.
gmpy has support for integer square root (returns the floor of the square root, calculated efficiently). This can be used for the is_square() function by testing if the square of the square root equals the original. ## Scheme code for fermat's Factorization

By : user3527560
Date : March 29 2020, 07:55 AM
I hope this helps you . Rather than using a while loop, in Scheme, you would just use recursion. Scheme has tail recursion, so recursion is just as performant as looping constructs in other languages.
Specifically, in this case, you would likely use something called "named let", which makes creating inline recursion easy. A direct translation of the above code to Scheme would result in this:
code :
``````(define (fermat-factor n)
(let* ((a (ceiling (sqrt n)))
(b (- (* a a) n)))
(let loop ()
(cond
((not (integer? (sqrt b)))
(set! a (+ a 1))
(set! b (- (* a a) n))
(loop))))
(- a (sqrt b))))
``````
``````(define (fermat-factor* n)
(let* ((a0 (ceiling (sqrt n)))
(b0 (- (* a0 a0) n)))
(let loop ((a a0)
(b b0))
(if (integer? (sqrt b))
(- a (sqrt b))
(loop (+ a 1)
(- (* a a) n))))))
`````` ## Largest Prime Factor Ruby using Fermat's factorization method

By : Md Anowar
Date : March 29 2020, 07:55 AM
I hope this helps . First note that Fermat factorisation doesn't give you prime factors in general.
Then, you run it until t+k >= n, that means you run the while loop n - t times, since t is roughly sqrt(n), that is an O(n) algorithm. For a largish n like 600851475143 (about 6*10^11), that is bound to take long. ## Why is this an incorrect implementation of Fermat's Factorization?

By : Sim
Date : March 29 2020, 07:55 AM
around this issue Typical issue with big number and floating points precision.
When you get to y = 323734167, you compute math.sqrt(n + y**2) which is math.sqrt(104804411734659032). 