## Pythrogorous Triplets

A Pythagorean triplet is a set of three integers a, b and c such that a2 + b2 = c2. Given a limit, generate all Pythagorean Triples with values smaller than given limit.

```Input : limit = 20
Output : 3 4 5
8 6 10
5 12 13
15 8 17
12 16 20```

Simple Solution is to generate these triplets smaller than given limit using three nested loop. For every triplet, check if Pythagorean condition is true, if true, then print the triplet. Time complexity of this solution is O(limit3) where ‘limit’ is given limit.

An Efficient Solution can print all triplets in O(k) time where k is number of triplets printed. The idea is to use square sum relation of Pythagorean triplet, i.e., addition of squares of a and b is equal to square of c, we can write these number in terms of m and n such that,

```       a = m2 - n2
b = 2 * m * n
c  = m2 + n2
because,
a2 = m4 + n4 – 2 * m2 * n2
b2 = 4 * m2 * n2
c2 = m4 + n4 + 2* m2 * n2
```

We can see that a2 + b2 = c2, so instead of iterating for a, b and c we can iterate for m and n and can generate these triplets.

Below is the implementation of above idea :

### C++

`// C++ program to generate pythagorean// triplets smaller than a given limit#include <bits/stdc++.h>// Function to generate pythagorean// triplets smaller than limitvoid pythagoreanTriplets(int limit){// triplet: a^2 + b^2 = c^2int a, b, c = 0;// loop from 2 to max_limititint m = 2;// Limiting c would limit// all a, b and cwhile (c < limit) {// now loop on j from 1 to i-1for (int n = 1; n < m; ++n) {// Evaluate and print triplets using// the relation between a, b and ca = m * m - n * n;b = 2 * m * n;c = m * m + n * n;if (c > limit)break;printf("%d %d %d\n", a, b, c);}m++;}}// Driver Codeint main(){int limit = 20;pythagoreanTriplets(limit);return 0;}`

### Java

`// Java program to generate pythagorean// triplets smaller than a given limitimport java.io.*;import java.util.*;class GFG {// Function to generate pythagorean// triplets smaller than limitstatic void pythagoreanTriplets(int limit){// triplet: a^2 + b^2 = c^2int a, b, c = 0;// loop from 2 to max_limititint m = 2;// Limiting c would limit// all a, b and cwhile (c < limit) {// now loop on j from 1 to i-1for (int n = 1; n < m; ++n) {// Evaluate and print// triplets using// the relation between// a, b and ca = m * m - n * n;b = 2 * m * n;c = m * m + n * n;if (c > limit)break;System.out.println(a + " " + b + " " + c);}m++;}}// Driver Codepublic static void main(String args[]){int limit = 20;pythagoreanTriplets(limit);}}`

### Python

`# Python3 program to generate pythagorean # triplets smaller than a given limit# Function to generate pythagorean # triplets smaller than limitdef pythagoreanTriplets(limits) :c, m = 0, 2# Limiting c would limit # all a, b and cwhile c < limits :# Now loop on n from 1 to m-1for n in range(1, m) :a = m * m - n * nb = 2 * m * nc = m * m + n * n# if c is greater than# limit then break itif c > limits :breakprint(a, b, c)m = m + 1# Driver Codeif __name__ == '__main__' :limit = 20pythagoreanTriplets(limit)`