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Fermat’s Factorization Method
  • Difficulty Level : Expert
  • Last Updated : 14 Jan, 2020

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

Example:

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.

Approach :

  1. If n = pq is a factorization of n into two positive integers, Then, since n is odd, so p and q are both odd.
  2. Let, a = 1/2 * (p+q) and b = 1/2 * (q-p).
  3. Since a and b are both integers, then p = (a – b) and q = (a + b).
  4. So, n = pq = (a – b)(a + b) = a2 – b2
  5. In case of prime number, we go back until b = 1 in as one factor is 1 for a prime number.
  6. A while loop ensures this operation

Below is the implementation of the above approach

C++




// C++ implementation of fermat's factorization 
#include<bits/stdc++.h>
  
using namespace std;
  
    // This function finds the value of a and b 
    // and returns a+b and a-b 
    void FermatFactors(int n) 
    
          
        // since fermat's factorization applicable 
        // for odd positive integers only 
        if(n <= 0) 
        
            cout << "[" << n << "]"
            return
        
      
        // check if n is a even number 
        if((n & 1) == 0) 
        
            cout << "[" << n / 2.0 << "," << 2 << "]"
            return
        
              
        int a = ceil(sqrt(n)) ; 
      
        // if n is a perfect root, 
        // then both its square roots are its factors 
        if(a * a == n) 
        
            cout << "[" << a << "," << a << "]"
            return
        
        int b; 
        while(true
        
            int b1 = a * a - n ; 
            b = (int)sqrt(b1) ; 
              
            if(b * b == b1) 
                break
            else
                a += 1; 
        
        cout << "[" << (a - b) << "," << (a + b) << "]"
        return
    
          
    // Driver Code 
    int main() 
    {
        FermatFactors(6557);
        return 0;
    }
  
// This code is contributed by AnkitRai01 


Java




// Java implementation of fermat's factorization 
class GFG 
{
      
    // This function finds the value of a and b 
    // and returns a+b and a-b 
    static void FermatFactors(int n)
    {
          
        // since fermat's factorization applicable 
        // for odd positive integers only 
        if(n <= 0)
        {
            System.out.print("["+ n + "]");
            return;
        }
      
        // check if n is a even number 
        if((n & 1) == 0)
        {
            System.out.print("[" + n / 2.0 + "," + 2 + "]"); 
            return;
        }
              
        int a = (int)Math.ceil(Math.sqrt(n)) ;
      
        // if n is a perfect root, 
        // then both its square roots are its factors 
        if(a * a == n)
        {
            System.out.print("[" + a + "," + a + "]"); 
            return;
        }
        int b;
        while(true)
        {
            int b1 = a * a - n ;
            b = (int)(Math.sqrt(b1)) ;
              
            if(b * b == b1)
                break;
            else
                a += 1;
        }
        System.out.print("[" + (a - b) +"," + (a + b) + "]" ); 
        return;
    }
          
    // Driver Code 
    public static void main (String[] args) 
    {
        FermatFactors(6557);
    }
}
  
// This code is contributed by AnkitRai01


Python3




# Python 3 implementation of fermat's factorization
  
from math import ceil, sqrt
  
#This function finds the value of a and b
#and  returns a+b and a-b
def FermatFactors(n):
  
   # since fermat's factorization applicable 
   # for odd positive integers only
    if(n<= 0):
        return [n]  
  
    # check if n is a even number 
    if(n & 1) == 0:  
        return [n / 2, 2
          
    a = ceil(sqrt(n))
  
    #if n is a perfect root, 
    #then both its square roots are its factors
    if(a * a == n):
        return [a, a]
  
    while(True):
        b1 = a * a -
        b = int(sqrt(b1))
        if(b * b == b1):
            break
        else:
            a += 1 
    return [a-b, a + b]
      
# Driver Code 
print(FermatFactors(6557))


C#




// C# implementation of fermat's factorization 
using System;
  
class GFG 
      
    // This function finds the value of a and b 
    // and returns a+b and a-b 
    static void FermatFactors(int n) 
    
          
        // since fermat's factorization applicable 
        // for odd positive integers only 
        if(n <= 0) 
        
            Console.Write("["+ n + "]"); 
            return
        
      
        // check if n is a even number 
        if((n & 1) == 0) 
        
            Console.Write("[" + n / 2.0 + "," + 2 + "]"); 
            return
        
              
        int a = (int)Math.Ceiling(Math.Sqrt(n)) ; 
      
        // if n is a perfect root, 
        // then both its square roots are its factors 
        if(a * a == n) 
        
            Console.Write("[" + a + "," + a + "]"); 
            return
        
        int b; 
        while(true
        
            int b1 = a * a - n ; 
            b = (int)(Math.Sqrt(b1)) ; 
              
            if(b * b == b1) 
                break
            else
                a += 1; 
        
        Console.Write("[" + (a - b) +"," + (a + b) + "]" ); 
        return
    
          
    // Driver Code 
    public static void Main () 
    
        FermatFactors(6557); 
    
  
// This code is contributed by AnkitRai01 


Output:

[79, 83]

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