Creating Sorting Method

We can solve this by sorting the array first.

1) Do square of every element in input array.

2) Sort the squared array in increasing order.

3) To find a triplet (a, b, c) such that a = b + c, do following.

1. Fix ‘a’ as last element of sorted array.
2. Now search for pair (b, c) in subarray between first element and ‘a’. A pair (b, c) with given sum can be found using meet in middle algorithm.
3. If no pair found for current ‘a’, then move ‘a’ one position back and repeat step 3.2.

// Returns true if there is a triplet with following property
// A[i]*A[i] = A[j]*A[j] + A[k]*[k]
// Note that this function modifies given array
static bool isTriplet(int[] arr, int n)
{
    // Square array elements
    for (int i = 0; i < n; i++)
        arr[i] = arr[i] * arr[i];

    // Sort array elements
    Array.Sort(arr);

    // If we reach here, then no triplet found
    return false;
}

Now we make 'a' as last element and search for pair between it and first element.

// Sort array elements
Arrays.sort(arr);

// Now fix one element one by one and find the other two
// elements
for (int i = n-1; i >= 2; i--)
{
    // To find the other two elements, start two index
    // variables from two corners of the array and move
    // them toward each other
    int l = 0; // index of the first element in arr[0..i-1]
    int r = i-1; // index of the last element in arr[0..i-1]
    while (l < r)
    {
        // A triplet found
        if (arr[l] + arr[r] == arr[i])
            return true;

        // Else either move 'l' or 'r'
        if (arr[l] + arr[r] < arr[i])
           l++;
        else
           r--;
    }
}