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# Euler’s Factorization method

• Last Updated : 12 Jul, 2021

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

• 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```
• 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)```
• 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.```
• 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```
• Therefore:
```(A + C) = M * H
(D + B) = L * H
where,
H = gcd((A + C), (D + B))```
• 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```
• Therefore,
`N = ((K/2)2 + (H/2)2)(L2 + M2)`
• Such that there exist 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```
• 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```
• 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 are even and divide them in half and find the factors.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement Eulers``// Factorization algorithm` `#include ``using` `namespace` `std;` `// Function to return N as the sum of``// two squares in two possible ways``void` `sumOfSquares(``int` `n, vector >& 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.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 > 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.first;``    ``b = vp.second;` `    ``c = vp.first;``    ``d = vp.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;``}`

## Java

 `// Java program to implement Eulers``// Factorization algorithm``import` `java.util.*;``class` `GFG{` `static` `class` `pair``{``  ``int` `first, second;``  ``public` `pair(``int` `first,``              ``int` `second) ``  ``{``    ``this``.first = first;``    ``this``.second = second;``  ``}   ``}``    ` `// Recursive function to``// return gcd of a and b ``static` `int` `__gcd(``int` `a, ``int` `b) ``{ ``  ``return` `b == ``0` `? a :``         ``__gcd(b, a % b);    ``}` `// Function to return N as the sum of``// two squares in two possible ways``static` `void` `sumOfSquares(``int` `n,``                         ``Vector vp)``{``  ``// Iterate a loop from 1 to Math.sqrt(n)``  ``for` `(``int` `i = ``1``; i <= Math.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 = (``int``)Math.sqrt(h);` `    ``// If h is perfect square``    ``if` `(h1 * h1 == h)``    ``{``      ``// Store in the sorted way``      ``int` `a = Math.max(h1, i),``          ``b = Math.min(h1, i);` `      ``// If there is already a pair``      ``// check if pairs are equal or not``      ``if` `(vp.size() == ``1` `&&``          ``a != vp.get(``0``).first)``        ``vp.add(``new` `pair(a, b));` `      ``// Insert the first pair``      ``if` `(vp.size() == ``0``)``        ``vp.add(``new` `pair(a, b));` `      ``// If two pairs are found``      ``if` `(vp.size() == ``2``)``        ``return``;``    ``}``  ``}``}` `// Function to find the factors``static` `void` `findFactors(``int` `n)``{``  ``// Get pairs where a^2 + b^2 = n``  ``Vector vp = ``new` `Vector<>();``  ``sumOfSquares(n, vp);` `  ``// Number cannot be represented``  ``// as sum of squares in two ways``  ``if` `(vp.size() != ``2``)``    ``System.out.print(``"Factors Not Possible"``);` `  ``// Assign a, b, c, d``  ``int` `a, b, c, d;` `  ``a = vp.get(``0``).first;``  ``b = vp.get(``0``).second;` `  ``c = vp.get(``1``).first;``  ``d = vp.get(``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``  ``System.out.print(``"a = "` `+ a +``                   ``"\t\t(A) a - c = "` `+ ``                   ``(a - c) +``                   ``"\t\tk = gcd[A, C] = "` `+``                   ``k + ``"\n"``);` `  ``System.out.print(``"b = "` `+ b +``                   ``"\t\t(B) a + c = "` `+ ``                   ``(a + c) +``                   ``"\t\th = gcd[B, D] = "` `+``                   ``h + ``"\n"``);` `  ``System.out.print(``"c = "` `+  c +``                   ``"\t\t(C) d - b = "` `+ ``                   ``(d - b) +``                   ``"\t\tl = A/k = "` `+``                   ``l + ``"\n"``);` `  ``System.out.print(``"d = "` `+  d +``                   ``"\t\t(D) d + b = "` `+``                   ``(d + b) +``                   ``"\t\tm = c/k = "` `+``                   ``m + ``"\n"``);` `  ``// Printing the factors``  ``if` `(k % ``2` `== ``0` `&& h % ``2` `== ``0``)``  ``{``    ``k = k / ``2``;``    ``h = h / ``2``;``    ``System.out.print(``"Factors are: "` `+``                     ``((k) * (k) + (h) * (h)) +``                     ``" "` `+ (l * l + m * m) + ``"\n"``);``  ``}``  ``else``  ``{``    ``l = l / ``2``;``    ``m = m / ``2``;``    ``System.out.print(``"Factors are: "` `+``                     ``((l) * (l) + (m) * (m)) +``                     ``" "` `+ (k * k + h * h) + ``"\n"``);``  ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``  ``int` `n = ``100000``;``  ``findFactors(n);``}``}` `// This code is contributed by gauravrajput1`

## C#

 `// C# program to implement Eulers``// Factorization algorithm``using` `System;``using` `System.Collections.Generic;``class` `GFG{` `public` `class` `pair``{``  ``public` `int` `first, second;``  ``public` `pair(``int` `first,``              ``int` `second) ``  ``{``    ``this``.first = first;``    ``this``.second = second;``  ``}   ``}``    ` `// Recursive function to``// return gcd of a and b ``static` `int` `__gcd(``int` `a, ``int` `b) ``{ ``  ``return` `b == 0 ? a :``         ``__gcd(b, a % b);    ``}` `// Function to return N as the sum of``// two squares in two possible ways``static` `void` `sumOfSquares(``int` `n,``                         ``List vp)``{``  ``// Iterate a loop from 1 to Math.Sqrt(n)``  ``for` `(``int` `i = 1; i <= Math.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 = (``int``)Math.Sqrt(h);` `    ``// If h is perfect square``    ``if` `(h1 * h1 == h)``    ``{``      ``// Store in the sorted way``      ``int` `a = Math.Max(h1, i),``          ``b = Math.Min(h1, i);` `      ``// If there is already a pair``      ``// check if pairs are equal or not``      ``if` `(vp.Count == 1 &&``          ``a != vp.first)``        ``vp.Add(``new` `pair(a, b));` `      ``// Insert the first pair``      ``if` `(vp.Count == 0)``        ``vp.Add(``new` `pair(a, b));` `      ``// If two pairs are found``      ``if` `(vp.Count == 2)``        ``return``;``    ``}``  ``}``}` `// Function to find the factors``static` `void` `findFactors(``int` `n)``{``  ``// Get pairs where a^2 + b^2 = n``  ``List vp = ``new` `List();``  ``sumOfSquares(n, vp);` `  ``// Number cannot be represented``  ``// as sum of squares in two ways``  ``if` `(vp.Count != 2)``    ``Console.Write(``"Factors Not Possible"``);` `  ``// Assign a, b, c, d``  ``int` `a, b, c, d;` `  ``a = vp.first;``  ``b = vp.second;` `  ``c = vp.first;``  ``d = vp.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``  ``Console.Write(``"a = "` `+ a +``                ``"\t\t(A) a - c = "` `+ ``                ``(a - c) +``                ``"\t\tk = gcd[A, C] = "` `+                   ``                ``k + ``"\n"``);` `  ``Console.Write(``"b = "` `+ b +``                ``"\t\t(B) a + c = "` `+ ``                ``(a + c) +``                ``"\t\th = gcd[B, D] = "` `+``                ``h + ``"\n"``);` `  ``Console.Write(``"c = "` `+  c +``                ``"\t\t(C) d - b = "` `+ ``                ``(d - b) +``                ``"\t\tl = A/k = "` `+``                ``l + ``"\n"``);` `  ``Console.Write(``"d = "` `+  d +``                ``"\t\t(D) d + b = "` `+``                ``(d + b) +``                ``"\t\tm = c/k = "` `+``                ``m + ``"\n"``);` `  ``// Printing the factors``  ``if` `(k % 2 == 0 && h % 2 == 0)``  ``{``    ``k = k / 2;``    ``h = h / 2;``    ``Console.Write(``"Factors are: "` `+``                  ``((k) * (k) + (h) * (h)) +``                  ``" "` `+ (l * l + m * m) + ``"\n"``);``  ``}``  ``else``  ``{``    ``l = l / 2;``    ``m = m / 2;``    ``Console.Write(``"Factors are: "` `+``                  ``((l) * (l) + (m) * (m)) +``                  ``" "` `+ (k * k + h * h) + ``"\n"``);``  ``}``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``  ``int` `n = 100000;``  ``findFactors(n);``}``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``
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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