Maximize value of (a+b) such that (a*a-b*b = N)

Given an odd value N, the task is to find the maximized value of (a+b) ( where a and b are integers ) such that (a² – b² = N)
Examples:

Input: N = 1
Output: 1
Since a*a - b*b = 1 
The maximum value occurs when a = 1 and b = 0. 
Thus, a + b = 1.

Input: N = 3
Output: 3
Since a*a - b*b = 3 
The maximum value occurs when a = 2 and b = 1. 
Thus, a + b = 3.

Approach:

• Given  

a*a - b*b = N
=> (a+b)*(a-b) = N
=> (a+b) = N/(a-b)

• Now for the above equation to be true  

We know that 
|a - b| ? 1
Therefore,
a + b ? N 

• Now inorder to maximise the value of (a + b), 
  

Maximising a + b ? N
=> a + b = N

• Hence, N is the maximum value of (a + b) when we need to maximize the value of (a+b) such that (a*a-b*b = N).
  

Below is the implementation of the above approach:

C++

// C++ program to maximize the value
// of (a+b) such that (a*a-b*b = N)

#include <bits/stdc++.h>
using namespace std;

// Function to maximize the value
// of (a+b) such that (a*a-b*b = n)
int maxValue(int n)
{
    return n;
}

// Driver code
int main()
{
    int n = 1;

    cout << maxValue(n);

    return 0;
}

Java

// Java program to maximize the value
// of (a+b) such that (a*a-b*b = N)
class GFG
{

// Function to maximize the value
// of (a+b) such that (a*a-b*b = n)
static int maxValue(int n)
{
    return n;
}

// Driver code
public static void main(String[] args)
{
    int n = 1;

    System.out.print(maxValue(n));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to maximize the value
# of (a+b) such that (a*a-b*b = N)

# Function to maximize the value
# of (a+b) such that (a*a-b*b = n)
def maxValue(n) :

    return n;

# Driver code
if __name__ == "__main__" :

    n = 1;

    print(maxValue(n));

# This code is contributed by AnkitRai01

C#

// C# program to maximize the value
// of (a+b) such that (a*a-b*b = N)
using System;

class GFG
{

    // Function to maximize the value
    // of (a+b) such that (a*a-b*b = n)
    static int maxValue(int n)
    {
        return n;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int n = 1;

        Console.Write(maxValue(n));
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript program to maximize the value
// of (a+b) such that (a*a-b*b = N)

// Function to maximize the value
// of (a+b) such that (a*a-b*b = n)
function maxValue(n)
{
    return n;
}

// Driver code
var n = 1;
document.write(maxValue(n));

</script>

Output:

1

Time Complexity:
Auxiliary Space: O(1)

As constant extra space is used.