# Count pairs (a, b) whose sum of squares is N (a^2 + b^2 = N)

Last Updated : 31 Aug, 2022

Given a number N, the task is to count all â€˜aâ€™ and â€˜bâ€™ that satisfy the condition a^2 + b^2 = N.
Note:- (a, b) and (b, a) are to be considered as two different pairs and (a, a) is also valid and to be considered only one time.
Examples:

```Input: N = 10
Output:  2
1^2 + 3^2 = 10
3^2 + 1^2 = 10

Input: N = 8
Output: 1
2^2 + 2^2 = 8```

Approach:

1. Traverse numbers from 1 to square root of N.
• Subtract square of the current number from N and check if their difference is a perfect square or not.
• If it is perfect square then increment the count.
2. Return count.

Below is the implementation of above approach:

## C++

 `// C++ program to count pairs whose sum` `// of squares is N` `#include ` `using` `namespace` `std;`   `// Function to count the pairs satisfying` `// a ^ 2 + b ^ 2 = N` `int` `countPairs(``int` `N)` `{` `    ``int` `count = 0;`   `    ``// Check for each number 1 to sqrt(N)` `    ``for` `(``int` `i = 1; i <= ``sqrt``(N); i++) {`   `        ``// Store square of a number` `        ``int` `sq = i * i;`   `        ``// Subtract the square from given N` `        ``int` `diff = N - sq;`   `        ``// Check if the difference is also` `        ``// a perfect square` `        ``int` `sqrtDiff = ``sqrt``(diff);`   `        ``// If yes, then increment count` `        ``if` `(sqrtDiff * sqrtDiff == diff)` `            ``count++;` `    ``}`   `    ``return` `count;` `}`   `// Driver code` `int` `main()` `{` `    ``// Loop to Count no. of pairs satisfying` `    ``// a ^ 2 + b ^ 2 = i for N = 1 to 10` `    ``for` `(``int` `i = 1; i <= 10; i++)` `        ``cout << ``"For n = "` `<< i << ``", "` `             ``<< countPairs(i) << ``" pair exists\n"``;`   `    ``return` `0;` `}`

## Java

 `// Java program to count pairs whose sum ` `// of squares is N `   `import` `java.io.*;`   `class` `GFG {`       `// Function to count the pairs satisfying ` `// a ^ 2 + b ^ 2 = N ` `static` `int` `countPairs(``int` `N) ` `{ ` `    ``int` `count = ``0``; `   `    ``// Check for each number 1 to sqrt(N) ` `    ``for` `(``int` `i = ``1``; i <= (``int``)Math.sqrt(N); i++) ` `    ``{ `   `        ``// Store square of a number ` `        ``int` `sq = i * i; `   `        ``// Subtract the square from given N ` `        ``int` `diff = N - sq; `   `        ``// Check if the difference is also ` `        ``// a perfect square ` `        ``int` `sqrtDiff = (``int``)Math.sqrt(diff); `   `        ``// If yes, then increment count ` `        ``if` `(sqrtDiff * sqrtDiff == diff) ` `            ``count++; ` `    ``} `   `    ``return` `count; ` `} `   `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)` `    ``{` `    ``// Loop to Count no. of pairs satisfying ` `    ``// a ^ 2 + b ^ 2 = i for N = 1 to 10 ` `    ``for` `(``int` `i = ``1``; i <= ``10``; i++) ` `        ``System.out.println( ``"For n = "` `+ i + ``", "` `            ``+ countPairs(i) + ``" pair exists\n"``); ` `    ``}` `}` `// This code is contributed by inder_verma.`

## Python 3

 `# Python 3 program to count pairs whose sum ` `# of squares is N `   `# From math import everything` `from` `math ``import` `*`   `# Function to count the pairs satisfying ` `# a ^ 2 + b ^ 2 = N ` `def` `countPairs(N) :` `    ``count ``=` `0`   `    ``# Check for each number 1 to sqrt(N) ` `    ``for` `i ``in` `range``(``1``, ``int``(sqrt(N)) ``+` `1``) :`   `        ``# Store square of a number ` `        ``sq ``=` `i ``*` `i`   `        ``# Subtract the square from given N` `        ``diff ``=` `N ``-` `sq`   `        ``#  Check if the difference is also ` `        ``# a perfect square ` `        ``sqrtDiff ``=` `int``(sqrt(diff))`   `        ``# If yes, then increment count` `        ``if` `sqrtDiff ``*` `sqrtDiff ``=``=` `diff :` `            ``count ``+``=` `1`   `    ``return` `count`   `# Driver code     ` `if` `__name__ ``=``=` `"__main__"` `:`   `    ``# Loop to Count no. of pairs satisfying ` `    ``# a ^ 2 + b ^ 2 = i for N = 1 to 10 ` `    ``for` `i ``in` `range``(``1``,``11``) :` `        ``print``(``"For n ="``,i,``", "``,countPairs(i),``"pair exists"``)`   ` `  `# This code is contributed by ANKITRAI1`

## C#

 `// C# program to count pairs whose sum ` `// of squares is N ` ` `    `using` `System; ` `class` `GFG {` ` `  ` `  ` `  `// Function to count the pairs satisfying ` `// a ^ 2 + b ^ 2 = N ` `static` `int` `countPairs(``int` `N) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``// Check for each number 1 to Sqrt(N) ` `    ``for` `(``int` `i = 1; i <= (``int``)Math.Sqrt(N); i++) ` `    ``{ ` ` `  `        ``// Store square of a number ` `        ``int` `sq = i * i; ` ` `  `        ``// Subtract the square from given N ` `        ``int` `diff = N - sq; ` ` `  `        ``// Check if the difference is also ` `        ``// a perfect square ` `        ``int` `sqrtDiff = (``int``)Math.Sqrt(diff); ` ` `  `        ``// If yes, then increment count ` `        ``if` `(sqrtDiff * sqrtDiff == diff) ` `            ``count++; ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main ()` `    ``{` `    ``// Loop to Count no. of pairs satisfying ` `    ``// a ^ 2 + b ^ 2 = i for N = 1 to 10 ` `    ``for` `(``int` `i = 1; i <= 10; i++) ` `        ``Console.Write( ``"For n = "` `+ i + ``", "` `            ``+ countPairs(i) + ``" pair exists\n"``); ` `    ``}` `}`

## PHP

 ``

## Javascript

 ``

Output:

```For n = 1, 1 pair exists
For n = 2, 1 pair exists
For n = 3, 0 pair exists
For n = 4, 1 pair exists
For n = 5, 2 pair exists
For n = 6, 0 pair exists
For n = 7, 0 pair exists
For n = 8, 1 pair exists
For n = 9, 1 pair exists
For n = 10, 2 pair exists```

Time Complexity : O(sqrt(N))

Auxiliary Space: O(1)

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