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+# Dimik - Square Number
+
+In this problem, you will be given `T` testcases. Each line of the testcase consists of an integer `n`. We just have to identify if the value of `n` is a square number or not.
+
+### Solution
+We can find the solution by square rooting the value of `n` using `sqrt` function and multiply against itself through the use of `floor` function. Because the value returned by `sqrt` function is `double` and to compare with the `integer` value of `n`, the datatype `double` is converted to `int` using `floor` function.
+
+### C++
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+int main()
+{
+    int t;
+    cin >> t;
+    for (int k = 1; k <= t; k++)
+    {
+        int n;
+        cin >> n;
+        if (floor(sqrt(n)) * floor(sqrt(n)) == n)
+            cout << "YES" << endl;
+        else
+            cout << "NO" << endl;
+    }
+}
+```