A non-negative integer is a perfect square when it equals some integer k multiplied by itself: n = k × k.
Remember
n is a perfect square ⇔ ∃ integer k ≥ 0 such that k² = n
Examples:
0 = 0²
1 = 1²
9 = 3²
16 = 4²
10 → no integer square root
n
√n idea
Perfect square?
16
4 × 4
Yes
25
5 × 5
Yes
10
between 3² and 4²
No
0
0 × 0
Yes
Approach
How to Check a Perfect Square in C#
Four clear steps — then pick an integer loop or a Sqrt-and-verify check.
Reject negatives (this tutorial treats only n >= 0).
Try candidate roots i = 0, 1, 2, … while i * i <= n.
If some i * i == n, return true; if you pass n, return false.
Or compute root = (int)Math.Sqrt(n) and verify root * root == n.
Method
Idea
Time
Best for
Integer loop
Compare i * i with n
O(√n)
Interviews without floating point
Sqrt + verify
Square the truncated root back
O(1)
Short production-style checks
Pseudocode
Pseudocode
function IsPerfectSquare(n):
if n < 0: return false
i = 0
while i * i <= n:
if i * i == n:
return true
i = i + 1
return false
Reference
Cheat Sheet
Keep these facts handy while writing the program.
Topic
Detail
Definition
n = k² for some integer k ≥ 0
Zero
0 is a perfect square
Negatives
Not perfect squares here
Loop stop
While i * i <= n
Sqrt verify
root * root == n after truncate
Classic checks
16 yes; 10 no; 25 yes
Try it
Live Preview
Enter an integer to see the nearest integer root and the perfect-square check.
Whole numbers only. Negatives are not perfect squares here. Tap a chip or type a value.
Result
n = 16
root = 4
4 × 4 = 16
16 is a perfect square
Trace
Worked Walkthrough — n = 16
Try successive integer squares until you match or pass 16.
Calculation
i = 0 → 0² = 0
i = 1 → 1² = 1
i = 2 → 2² = 4
i = 3 → 3² = 9
i = 4 → 4² = 16 ← equal to n
16 is a perfect square
Sqrt path:
Math.Sqrt(16) = 4.0
root = 4
4 * 4 == 16 → same answer
Code
C# Programs
Start with the integer loop so you never rely on floating point. Use Sqrt-and-verify when you want a short, clear check.
Example 1 — Integer Loop
Increase i while i * i <= n and look for an exact match.
C#
using System;
class Program
{
static bool IsPerfectSquare(int n)
{
if (n < 0)
{
return false;
}
for (int i = 0; i * i <= n; i++)
{
if (i * i == n)
{
return true;
}
}
return false;
}
static void Main()
{
int number = 16;
Console.WriteLine(number + (IsPerfectSquare(number) ? " is a perfect square" : " is not a perfect square"));
}
}
Output
16 is a perfect square
How It Works
Negatives return false immediately.
The loop condition i * i <= n stops once squares go past n.
Each step checks whether i * i equals n.
For 16, i = 4 matches and the method returns true.
If no match appears before the loop ends, the answer is false.
Example 2 — Sqrt + Verify
Take the square root, truncate to an integer, then square it back.
C#
using System;
class Program
{
static bool IsPerfectSquare(int n)
{
if (n < 0)
{
return false;
}
int root = (int)Math.Sqrt(n);
return root * root == n;
}
static void Main()
{
Console.WriteLine(IsPerfectSquare(16)); // True
Console.WriteLine(IsPerfectSquare(10)); // False
Console.WriteLine(IsPerfectSquare(0)); // True
Console.WriteLine(IsPerfectSquare(-4)); // False
}
}
Output
True
False
True
False
How It Works
Same non-negative rule as Example 1.
Math.Sqrt(n) returns a double approximation of √n.
Casting to int keeps the integer part of that root.
Squaring root again and comparing with n confirms an exact square.
For 10, root becomes 3 and 9 != 10, so the result is false.
Edge Cases & Pitfalls
Check these before calling the solution done.
0
Zero counts
0 = 0². Do not special-case it as false.
Negatives
Not squares here
Return false for n < 0 in this beginner definition.
Overflow
i * i care
For large n, use long for i * i so the product does not overflow int.
Sqrt
Always verify
Never decide from the double alone — square the truncated root back.
Analysis
Time and Space Complexity
Program
Time
Extra space
Integer loop
O(√n)
O(1)
Sqrt + verify
O(1)
O(1)
Both are fine for interviews. Prefer the loop when you must avoid floating-point discussion.
Remember
Key Takeaways
Definition:n = k² for some integer k ≥ 0.
Loop: test i * i while i * i <= n.
Sqrt: truncate, then verify root * root == n.
Test set: 0 → yes; 16 → yes; 10 → no; -4 → no.
One line: find an integer root whose square equals n — by looping or by verifying Math.Sqrt.
Frequently Asked Questions
A non-negative integer that equals some integer multiplied by itself. Examples: 0, 1, 4, 9, 16, 25.
Loop with i*i until you match or pass n, or take Math.Sqrt(n), cast to int, and verify root*root == n.
Yes. 0 = 0 × 0.
Sqrt returns a double. Floating-point rounding can misclassify edge values, so always square the integer root back and compare.
A loop up to √n is O(√n). The Sqrt-and-verify method is O(1) for practical interview sizes (constant-time math call plus a few ops).
