Fermat’s Factorization Method

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 - n
        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

Javascript

<script>

    // JavaScript implementation of fermat's factorization

    // This function finds the value of a and b
    // and returns a+b and a-b
    function FermatFactors(n)
    {

        // since fermat's factorization applicable
        // for odd positive integers only
        if(n <= 0)
        {
            document.write("["+ n + "]");
            return;
        }

        // check if n is a even number
        if((n & 1) == 0)
        {
            document.write("[" + (n / 2.0) + "," + (2) + "]");
            return;
        }

        let a = Math.ceil(Math.sqrt(n)) ;

        // if n is a perfect root,
        // then both its square roots are its factors
        if(a * a == n)
        {
            document.write("[" + a + "," + a + "]");
            return;
        }
        let b;
        while(true)
        {
            let b1 = a * a - n ;
            b = parseInt(Math.sqrt(b1), 10);

            if(b * b == b1)
                break;
            else
                a += 1;
        }
        document.write("[" + (a - b) +", " + (a + b) + "]");
        return;
    }

    FermatFactors(6557);

</script>

Output:

[79, 83]

Time Complexity :   O(sqrt(n))

Space Complexity : O(1)