Euler’s Factorization method

Given a number N, the task is to find the factors of N.

Examples:

Input: N = 1000009
Output: 293 3413
Explanation:
293 * 3413 = 1000009

Input: N = 100000
Output: 800 125
Explanation:
800 * 125 = 100000

Euler’s Factorization method: Euler’s factorization method works on the principle that all the numbers N which can be written as the sum of two powers in two different ways can be factored into two numbers, (i.e) N = A2 + B2 = C2 + D2 where A != C and A != D, then there exists two factors for N.



Working of the algorithm: Let N be the number for which we need to find the factors.

  1. So, initially, we need to find two ways to represent N as the sum of powers of two numbers.
    N = A2 + B2
    N = C2 + D2
    Therefore,
    N = A2 + B2 = C2 + D2
    
  2. Now, the algebraic operations are performed on the above equation to convert the equations as:
    N = A2 + B2 = C2 + D2
    -> N = A2 - C2 = D2 - B2
    -> N = (A - C)(A + C) = (D - B)(D + B)
    
  3. Let K be the GCD of (A – C) and (D – B). So,
    A - C = K * L
    D - B = K * M
    where GCD(L, M) is 1.
    
  4. Clearly, L = (A – C) / K and M = (D – B)/K. On substituting this in the initial equation:
    N = K * L * (A + C) = K * M * (D + B)
    -> L * (A + C) = M * (D + B)
    -&gtl (A + C)/(D + B) = M/L
    
  5. Therefore:
    (A + C) = M * H 
    (D + B) = L * H
    where,
    H = gcd((A + C), (D + B))
    
  6. Let Q = (K2 + H2)(L2 + M2).
    -> ((KL)2 + (KM)2 + (HL)2 + (HM)2)
    -> ((A - C)2 + (D - B)2 + (D + B)2 + (A + C)2)
    -> ((2 * A)2 + (2 * B)2 + (2 * C)2 + (2 * D)2)
    -> 4 * N
    
  7. Therefore,
    N = ((K/2)2 + (H/2)2)(L2 + M2)
    

    Such that there exists a pair K and H which are both even numbers.

Let’s visualize the above approach by taking an example. Let N = 221.

  1. 221 = 112 + 102 = 52 + 142
  2. From the above equation:
    A = 11 - 5 = 6
    B = 11 + 5 = 15
    C = 14 - 10 = 4
    D = 14 + 10 = 24
    
  3. Therefore, the above values can be used to compute the values of K, H, L and M.
    K = GCD(6, 4) = 2
    H = GCD(16, 24) = 8
    L = GCD(6, 24) = 3
    M = GCD(16, 4) = 2
    
  4. Therefore:
    221 = ((2/2)2 + (8/2)2) * (32 + 22)
    221 = 17 * 13
    

Approach: In order to implement the above approach, the following steps are computed:

  1. Find the sum of squares by iterating a loop from 1 to sqrt(N) because no factor exists between [sqrt(N), N] apart from N and find two pairs whose sum of squares is equal to N.
  2. Store the values in A, B, C, D.
  3. Find the values of K, H, L and M using the formula mentioned in the above approach.
  4. Use the values of K, H, L and M to find the factors. Check the pair where both the numbers is even and divide them in half and find the factors.

Below is the implementation of the above approach:

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// C++ program to implement Eulers
// Factorization algorithm
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to return N as the sum of
// two squares in two possible ways
void sumOfSquares(int n, vector<pair<int, int> >& vp)
{
    // Iterate a loop from 1 to sqrt(n)
    for (int i = 1; i <= sqrt(n); i++) {
  
        // If i*i is square check if there
        // exists another integer such that
        // h is a perfect square and i*i + h = n
        int h = n - i * i, h1 = sqrt(h);
  
        // If h is perfect square
        if (h1 * h1 == h) {
  
            // Store in the sorted way
            int a = max(h1, i), b = min(h1, i);
  
            // If there is already a pair
            // check if pairs are equal or not
            if (vp.size() == 1 && a != vp[0].first)
                vp.push_back(make_pair(a, b));
  
            // Insert the first pair
            if (vp.size() == 0)
                vp.push_back(make_pair(a, b));
  
            // If two pairs are found
            if (vp.size() == 2)
                return;
        }
    }
}
  
// Function to find the factors
void findFactors(int n)
{
  
    // Get pairs where a^2 + b^2 = n
    vector<pair<int, int> > vp;
    sumOfSquares(n, vp);
  
    // Number cannot be represented
    // as sum of squares in two ways
    if (vp.size() != 2)
        cout << "Factors Not Possible";
  
    // Assign a, b, c, d
    int a, b, c, d;
  
    a = vp[0].first;
    b = vp[0].second;
  
    c = vp[1].first;
    d = vp[1].second;
  
    // Swap if a < c because
    // if a - c < 0,
    // GCD cant be computed.
    if (a < c) {
        int t = a;
        a = c;
        c = t;
        t = b;
        b = d;
        d = t;
    }
  
    // Compute the values of k, h, l, m
    // using the formula mentioned
    // in the approach
    int k, h, l, m;
    k = __gcd(a - c, d - b);
    h = __gcd(a + c, d + b);
    l = (a - c) / k;
    m = (d - b) / k;
  
    // Print the values of a, b, c, d
    // and k, l, m, h
    cout << "a = " << a
         << "\t\t(A) a - c = " << (a - c)
         << "\t\tk = gcd[A, C] = "
         << k << endl;
  
    cout << "b = " << b
         << "\t\t(B) a + c = " << (a + c)
         << "\t\th = gcd[B, D] = "
         << h << endl;
  
    cout << "c = " << c
         << "\t\t(C) d - b = " << (d - b)
         << "\t\tl = A/k = "
         << l << endl;
  
    cout << "d = " << d
         << "\t\t(D) d + b = " << (d + b)
         << "\t\tm = c/k = "
         << m << endl;
  
    // Printing the factors
    if (k % 2 == 0 && h % 2 == 0) {
        k = k / 2;
        h = h / 2;
  
        cout << "Factors are: "
             << ((k) * (k) + (h) * (h))
             << " " << (l * l + m * m)
             << endl;
    }
    else {
        l = l / 2;
        m = m / 2;
  
        cout << "Factors are: "
             << ((l) * (l) + (m) * (m))
             << " " << (k * k + h * h)
             << endl;
    }
}
  
// Driver code
int main()
{
    int n = 100000;
  
    findFactors(n);
  
    return 0;
}

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Output:

a = 316        (A) a - c = 16        k = gcd[A, C] = 8
b = 12        (B) a + c = 616        h = gcd[B, D] = 56
c = 300        (C) d - b = 88        l = A/k = 2
d = 100        (D) d + b = 112        m = c/k = 11
Factors are: 800 125

Complexity Analysis:
Time Complexity: O(sqrt(N)), where N is the given number
Space Complexity: O(1)

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