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Pair of integers (a, b) which satisfy the given equations

  • Last Updated : 30 Jan, 2022

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:
Explanation: 
Only one pair (3, 0) exists for both equations satisfying the conditions.
Input: n = 4, m = 20 
Output:
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 pairs
int 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 code
int 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 = m
class GFG
{
 
// Function to count valid pairs
static 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 code
public 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 pairs
def 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 code
if __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 = m
using System;
 
class GFG
{
 
// Function to count valid pairs
static 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 code
public 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 pairs
function 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 code
var 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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