Fermat’s Factorization method for large numbers

Given a large number N, the task is to divide this number into a product of two factors, using Fermat’s Factorisation method. Examples

Input: N = 105327569 Output: 10223, 10303 Input: N = 249803 Output: 23, 10861

Fermat Factorization: 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.

Approach:

1. Get the number as an object of BigInteger class
2. Find the square root of N.
3. It is guaranteed that the value of a is greater than sqrt(N) and value of b less than sqrt(N).
4. Take the value of sqrt(n) as a and increment the number until and unless a number b is found such that N – a^2 is a perfect square.

Below is the implementation of the above approach: 

Java

// Java program for Fermat's Factorization
// method for large numbers

import java.math.*;
import java.util.*;

class Solution {

    // Function to find the Floor
    // of square root of a number
    public static BigInteger sqrtF(BigInteger x)
        throws IllegalArgumentException
    {
        // if x is less than 0
        if (x.compareTo(BigInteger.ZERO) < 0) {
            throw new IllegalArgumentException(
                "Negative argument.");
        }

        // if x==0 or x==1
        if (x.equals(BigInteger.ZERO)
            || x.equals(BigInteger.ONE)) {
            return x;
        }

        BigInteger two
            = BigInteger.valueOf(2L);
        BigInteger y;

        // run a loop
        y = x.divide(two);
        while (y.compareTo(x.divide(y)) > 0)
            y = ((x.divide(y)).add(y))
                    .divide(two);
        return y;
    }

    // function to find the Ceil
    // of square root of a number
    public static BigInteger sqrtC(BigInteger x)
        throws IllegalArgumentException
    {
        BigInteger y = sqrtF(x);

        if (x.compareTo(y.multiply(y)) == 0) {
            return y;
        }

        else {
            return y.add(BigInteger.ONE);
        }
    }

    // Fermat factorisation
    static String FermatFactors(BigInteger n)
    {
        // constants
        BigInteger ONE = new BigInteger("1");
        BigInteger ZERO = new BigInteger("0");
        BigInteger TWO = new BigInteger("2");

        // if n%2 ==0 then return the factors
        if (n.mod(TWO).equals(ZERO)) {
            return n.divide(TWO)
                       .toString()
                + ", 2";
        }

        // find the square root
        BigInteger a = sqrtC(n);

        // if the number is a perfect square
        if (a.multiply(a).equals(n)) {
            return a.toString()
                + ", " + a.toString();
        }

        // else perform factorisation
        BigInteger b;
        while (true) {
            BigInteger b1 = a.multiply(a)
                                .subtract(n);
            b = sqrtF(b1);

            if (b.multiply(b).equals(b1))
                break;
            else
                a = a.add(ONE);
        }

        return a.subtract(b).toString()
            + ", " + a.add(b).toString();
    }

    // Driver code
    public static void main(String args[])
    {
        String N = "105327569";

        System.out.println(
            FermatFactors(
                new BigInteger(N)));
    }
}

Python3

import math

# Function to find the Floor
# of square root of a number
def sqrtF(x):
    # if x is less than 0
    if x < 0:
        raise ValueError("Negative argument.")

    # if x==0 or x==1
    if x == 0 or x == 1:
        return x

    y = x // 2

    # run a loop
    while y > x // y:
        y = (x // y + y) // 2

    return y

# function to find the Ceil
# of square root of a number
def sqrtC(x):
    y = sqrtF(x)

    if x == y * y:
        return y
    else:
        return y + 1

# Fermat factorisation
def FermatFactors(n):
    # if n%2 ==0 then return the factors
    if n % 2 == 0:
        return str(n // 2) + ", 2"

    # find the square root
    a = sqrtC(n)

    # if the number is a perfect square
    if a * a == n:
        return str(a) + ", " + str(a)

    # else perform factorisation
    while True:
        b1 = a * a - n
        b = sqrtF(b1)

        if b * b == b1:
            break
        else:
            a += 1

    return str(a - b) + ", " + str(a + b)

# Driver code
if __name__ == "__main__":
    N = "105327569"

    print(FermatFactors(int(N)))
# This code is contributed by sarojmcy2e

Javascript

function sqrtF(x) {
  if (x < 0) {
    throw new Error("Negative argument.");
  }

  if (x === 0 || x === 1) {
    return x;
  }

  let y = BigInt(x >> 1n);
  while (y > BigInt(x / y)) {
    y = BigInt((BigInt(x) / BigInt(y) + BigInt(y)) >> 1n);
  }
  return y;
}

function sqrtC(x) {
  const y = sqrtF(x);

  if (x === y * y) {
    return y;
  } else {
    return y + 1n;
  }
}

function FermatFactors(n) {
  const ONE = 1n;
  const ZERO = 0n;
  const TWO = 2n;

  if (n % TWO === ZERO) {
    return `${n / TWO}, 2`;
  }

  const a = sqrtC(n);

  if (a * a === n) {
    return `${a}, ${a}`;
  }

  let b;
  while (true) {
    const b1 = a * a - n;
    b = sqrtF(b1);

    if (b * b === b1) {
      break;
    } else {
      a += 1n;
    }
  }

  return `${a - b}, ${a + b}`;
}

const N = "105327569";
console.log(FermatFactors(BigInt(N)));

Output:

10223, 10303

Performance Analysis : 

Time Complexity: O(sqrt(N))
Space Complexity: O(1)