Check if the square of a number is divisible by K or not

Given two integers, X and K, the task is to find if X² is divisible by K or not. Here, both K and X can lie in the range [1,10¹⁸].

Examples:

Input: X = 6, K = 9 
Output: YES 
Explanation:
Since 6² is equal to 36, which is divisible by 9.

Input: X = 7, K = 21 
Output: NO 
Explanation:
Since 7² is equal to 49, which is not divisible by 21. 

Approach:
As mentioned above, X can lie in the range [1,10¹⁸], so we can not directly check if X² is divisible by K or not as it can cause a memory overflow. Hence the efficient approach is to first calculate the Greatest Common Divisor of X and K. After the GCD is calculated, we will take it as the maximum portion which we can divide from X and K. Reduce K by GCD and check if X is divisible by the reduced K or not.

Below is the implementation of above approach: 

C++

// C++ implementation to
// check if the square of X is
// divisible by K

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

// Function to return if
// square of X is divisible
// by K
void checkDivisible(int x, int k)
{
    // Finding gcd of x and k
    int g = __gcd(x, k);

    // Dividing k by their gcd
    k /= g;

    // Check for divisibility of X
    // by reduced K
    if (x % k == 0) {
        cout << "YES\n";
    }
    else {
        cout << "NO\n";
    }
}

// Driver Code
int main()
{
    int x = 6, k = 9;
    checkDivisible(x, k);

    return 0;
}

Java

// Java implementation to
// check if the square of X is
// divisible by K

class GFG{

// Function to return if
// square of X is divisible
// by K
static void checkDivisible(int x, int k)
{
    // Finding gcd of x and k
    int g = __gcd(x, k);

    // Dividing k by their gcd
    k /= g;

    // Check for divisibility of X
    // by reduced K
    if (x % k == 0)
    {
        System.out.print("YES\n");
    }
    else
    {
        System.out.print("NO\n");
    }
}
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);
}

// Driver Code
public static void main(String[] args)
{
    int x = 6, k = 9;
    checkDivisible(x, k);
}
}

// This code is contributed by gauravrajput1

Python3

# Python3 implementation to
# check if the square of X is
# divisible by K

from math import gcd

# Function to return if
# square of X is divisible
# by K
def checkDivisible(x, k):

    # Finding gcd of x and k
    g = gcd(x, k)

    # Dividing k by their gcd
    k //= g

    # Check for divisibility of X
    # by reduced K
    if (x % k == 0):
        print("YES")
    else:
        print("NO")

# Driver Code
if __name__ == '__main__':

    x = 6
    k = 9
    checkDivisible(x, k);

# This code is contributed by Bhupendra_Singh

C#

// C# implementation to check
// if the square of X is
// divisible by K
using System;

class GFG{

// Function to return if
// square of X is divisible
// by K
static void checkDivisible(int x, int k)
{

    // Finding gcd of x and k
    int g = __gcd(x, k);

    // Dividing k by their gcd
    k /= g;

    // Check for divisibility of X
    // by reduced K
    if (x % k == 0)
    {
        Console.Write("YES\n");
    }
    else
    {
        Console.Write("NO\n");
    }
}

static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);
}

// Driver Code
public static void Main(String[] args)
{
    int x = 6, k = 9;

    checkDivisible(x, k);
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// Javascript implementation to
// check if the square of X is
// divisible by K

// Return gcd of two numbers
function gcd(a, b)
{
    return b == 0 ? a : gcd(b, a % b);
}

// Function to return if
// square of X is divisible
// by K
function checkDivisible(x, k)
{

    // Finding gcd of x and k
    var g = gcd(x, k);

    // Dividing k by their gcd
    k /= g;

    // Check for divisibility of X
    // by reduced K
    if (x % k == 0)
    {
        document.write("YES");
    }
    else
    {
        document.write("NO");
    }
}

// Driver code
var x = 6, k = 9;
checkDivisible(x, k);

// This code is contributed by Ankita saini

</script>

Output:

YES

Time Complexity: O(log(max(x, k)))

Auxiliary Space;:O(1)