* Author:* @somashekhar001

*@somashekhar001*

**Tester:***@somashekhar001*

**Editorialist:**# DIFFICULTY:

SIMPLE

# PREREQUISITES:

Mathematics

# PROBLEM:

(a,b,c) is a pythagorean triplet you are given b and b is even and you have to find a and c and output maximum sum of triplet (a+b+c),

# EXPLANATION:

Let us consider n as a number.

```
4 * n^2 = 4 * n^2 ( multiply 4 both side)
= ( n^2 + 1 )^2 - ( n^2 - 1 )^2
( n^2 - 1 )^2 + ( 2 * n )^2 = ( n^2 + 1 )^2
a^2 + b^2 = c^2
a = n^2 - 1
b = 2 * n
c= n^2 + 1
sum=a+b+c;
=n^2 - 1 + b + n^2 + 1
=2*n^2+b (n=b/2)
=(b^2 / 2 )+b
```

# SOLUTIONS:

#include<bits/stdc++.h>

using namespace std;

int main()

{

long long T;

cin>>T;//input T testcases

while(Tâ€“)

{

long long b;

cin>>b;//taking input b which is even

cout<<((b*b)/2)+b<<endl;

}

return 0;

}