Given the system of equations a2 + b = n and a + b2 = m. The task is to find the number of pair of positive integers (a, b) which satisfy the equation for given n and m.
Examples:
Input: n = 9, m = 3
Output: 1
Explanation:
Only one pair (3, 0) exists for both equations satisfying the conditions.
Input: n = 4, m = 20
Output: 0
Explanation:
There are no such pair exists.
Approach:
The approach is to check for all possible pairs of numbers and check if that pair satisfy both the equations or not. For this we have,
a2 + b = n ... (1) a + b2 = m ... (2) For equation (2), => a = m - b2 ... (3)
- Now for the positive value of a, every value of b must be from 0 to sqrt(m).
- Obtain the value of a from equations (3).
- If the pair (a, b) satisfy equation (1), then pair (a, b) is the solution of system of equations.
Below is the implementation of the above approach:
C++
// C++ program to count the pair of integers(a, b)// which satisfy the equation// a^2 + b = n and a + b^2 = m#include <bits/stdc++.h>using namespace std;// Function to count valid pairsint pairCount(int n, int m){ int cnt = 0, b, a; for (b = 0; b <= sqrt(m); b++) { a = m - b * b; if (a * a + b == n) { cnt++; } } return cnt;}// Driver codeint main(){ int n = 9, m = 3; cout << pairCount(n, m) << endl; return 0;} |
Java
// Java program to count the pair of integers(a, b)// which satisfy the equation// a^2 + b = n and a + b^2 = mclass GFG{// Function to count valid pairsstatic int pairCount(int n, int m){ int cnt = 0, b, a; for (b = 0; b <= Math.sqrt(m); b++) { a = m - b * b; if (a * a + b == n) { cnt++; } } return cnt;}// Driver codepublic static void main(String[] args){ int n = 9, m = 3; System.out.print(pairCount(n, m) +"\n");}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to count the pair of integers(a, b)# which satisfy the equation# a^2 + b = n and a + b^2 = m# Function to count valid pairsdef pairCount(n, m): cnt = 0; for b in range(int(pow(m, 1/2))): a = m - b * b; if (a * a + b == n): cnt += 1; return cnt;# Driver codeif __name__ == '__main__': n = 9; m = 3; print(pairCount(n, m)); # This code is contributed by 29AjayKumar |
C#
// C# program to count the pair of integers(a, b)// which satisfy the equation// a^2 + b = n and a + b^2 = musing System;class GFG{// Function to count valid pairsstatic int pairCount(int n, int m){ int cnt = 0, b, a; for (b = 0; b <= Math.Sqrt(m); b++) { a = m - b * b; if (a * a + b == n) { cnt++; } } return cnt;}// Driver codepublic static void Main(String[] args){ int n = 9, m = 3; Console.Write(pairCount(n, m) +"\n");}}// This code is contributed by 29AjayKumar |
Javascript
<script>// javascript program to count the pair of integers(a, b)// which satisfy the equation// a^2 + b = n and a + b^2 = m// Function to count valid pairsfunction pairCount(n , m){ var cnt = 0, b, a; for (b = 0; b <= Math.sqrt(m); b++) { a = m - b * b; if (a * a + b == n) { cnt++; } } return cnt;}// Driver codevar n = 9, m = 3;document.write(pairCount(n, m));// This code is contributed by Amit Katiyar </script> |
Output:
1
Time Complexity: O(sqrt(min(n,m))
Auxiliary Space: O(1)
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