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### Problem
Given two numbers `n` and  `k`, the task is to check whether the given number `n` can be made a perfect square after adding `k` to it.

### Examples
* Input: `n = 10`, `k = 6`

* Output: `YES`

Explanation: $10 + 6 = 16$, which is a perfect square of $4$.

* Input: `n = 11`, `k = 8`

* Output: `NO`

Explanation: $11 + 8 = 19$, which is a not perfect square.


### Solution
 * Add `n` and `k` (`n+k`), and then find the square root for `n+k`
* Check if `num_sqrt`*`num_sqrt`  = `n+k`
* Print `YES` if it is a perfect square, otherwise print `NO`

### Implementation in Python
In the following code snippet, we:
* In line 1: Importing `sqrt` from module `math` to calculate square root.
* In line 3 to 4: Initialize `n` and `k` with `10` and `6` respectively.
* In line 6: Add `k` to `n` and store in `num`.
* In line 8: Calculating the square root of `num`.
* In line 10 to 13: Check for a perfect square.

It will print `YES` as $10 + 6 = 16$, square root $16$ is $4$, and satisfies the condition for perfect square as $4 * 4 = 16$.

from math import sqrt

#initialize N and K
N = 10
K = 6

#add K to N
num = N+K

#calculate square root
num_sqrt = int(sqrt(num))

#check for perfect square
if(num_sqrt * num_sqrt == num):
  print("YES")
else:
  print("NO")

The following code snippet also follows the same procedure as above except we changed the values for `n` and `k`.

It will print `NO` as $11 + 8 = 19$, which is not a perfect square. If we calculate square root for $19$, we get $4$, which doesn't satisfy the condition for perfect square. $4 * 4 = 16$, not $19$.

from math import sqrt

#initialize N and K
N = 11
K = 8

#add K to N
num = N+K

#calculate square root
num_sqrt = int(sqrt(num))

#check for perfect square
if(num_sqrt * num_sqrt == num):
  print("YES")
else:
  print("NO")

How to check whether a number can be made a perfect square

Problem

Examples

Solution

Implementation in Python