Check if given number is perfect square in Python

A perfect square is a number that can be expressed as the product of an integer with itself. In other words, a number is a perfect square when its square root is an integer. For example, 36 is a perfect square because ?36 = 6, and 6 is an integer.

In this tutorial, we'll explore different methods to check if a given number is a perfect square in Python.

Algorithm

To check if a number is a perfect square, we follow these steps ?

• Calculate the square root of the given number
• Take the integer part of the square root
• Square this integer and compare it with the original number
• If they match, the number is a perfect square

Method 1: Using math.sqrt()

The most straightforward approach uses Python's built-in math.sqrt() function ?

from math import sqrt

def is_perfect_square(n):
    if n < 0:
        return False
    
    sq_root = int(sqrt(n))
    return (sq_root * sq_root) == n

# Test with different numbers
test_numbers = [36, 25, 10, 16, 0, 1]

for num in test_numbers:
    result = is_perfect_square(num)
    print(f"{num} is perfect square: {result}")

36 is perfect square: True
25 is perfect square: True
10 is perfect square: False
16 is perfect square: True
0 is perfect square: True
1 is perfect square: True

Method 2: Using Integer Square Root

For better precision with large numbers, we can use the integer square root method ?

def is_perfect_square_int(n):
    if n < 0:
        return False
    
    # Using integer square root to avoid floating point errors
    x = n
    while True:
        y = (x + n // x) // 2
        if y >= x:
            break
        x = y
    
    return x * x == n

# Test the function
numbers = [144, 145, 169, 200]

for num in numbers:
    result = is_perfect_square_int(num)
    print(f"{num} is perfect square: {result}")

144 is perfect square: True
145 is perfect square: False
169 is perfect square: True
200 is perfect square: False

Method 3: Using math.isqrt() (Python 3.8+)

Python 3.8 introduced math.isqrt() which returns the integer square root ?

import math

def is_perfect_square_isqrt(n):
    if n < 0:
        return False
    
    sq_root = math.isqrt(n)
    return sq_root * sq_root == n

# Test with various numbers
test_cases = [49, 50, 64, 100, 121]

for num in test_cases:
    result = is_perfect_square_isqrt(num)
    print(f"?{num} = {math.isqrt(num)}, Perfect square: {result}")

?49 = 7, Perfect square: True
?50 = 7, Perfect square: False
?64 = 8, Perfect square: True
?100 = 10, Perfect square: True
?121 = 11, Perfect square: True

Comparison of Methods

Conclusion

Use math.isqrt() for Python 3.8+ as it provides perfect integer precision. For older versions, math.sqrt() works well for most cases, while the integer square root method handles large numbers more reliably.