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

Namachila November 19, 2025

0 0 3 minutes read

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)

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