Fermat’s Factorization method is based on the representation of an odd integer as the difference of two squares.
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
Input: n = 6557 Output: [79,83] Explanation: For the above value, the first try for a is ceil value of square root of 6557, which is 81. Then, b2 = 812 - 6557 = 4, as it is a perfect square. So, b = 2 So, the factors of 6557 are: (a - b) = 81 -2 = 79 & (a + b) = 81 + 2 = 83.
- If n = pq is a factorization of n into two positive integers, Then, since n is odd, so p and q are both odd.
- Let, a = 1/2 * (p+q) and b = 1/2 * (q-p).
- Since a and b are both integers, then p = (a – b) and q = (a + b).
- So, n = pq = (a – b)(a + b) = a2 – b2
- In case of prime number, we go back until b = 1 in as one factor is 1 for a prime number.
- A while loop ensures this operation
Below is the implementation of the above approach
- Euler's Factorization method
- Dixon's Factorization Method with implementation
- Fermat's Factorization method for large numbers
- Wheel Factorization Algorithm
- Pollard's Rho Algorithm for Prime Factorization
- Sum of Factors of a Number using Prime Factorization
- Trial division Algorithm for Prime Factorization
- Prime Factorization using Sieve O(log n) for multiple queries
- Count occurrences of a prime number in the prime factorization of every element from the given range
- Transportation Problem | Set 6 (MODI Method - UV Method)
- Program for Bisection Method
- Program for Muller Method
- Java Math incrementExact(int x) method
- Program for Newton Raphson Method
- Primality Test | Set 2 (Fermat Method)
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Improved By : AnkitRai01