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

• Last Updated : 13 Apr, 2021

Given a number N, the task is to count all ‘a’ and ‘b’ 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 = 9
3^2 + 1^2 = 9

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))

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