How to Check if a Given Number is Fibonacci number Python (original) (raw)

Last Updated : 23 Jul, 2025

Fibonacci numbers are part of a famous sequence where each number is the sum of the two preceding ones, i.e. **F(n) = F(n-1) + F(n-2). The sequence starts as:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

Notice that every number is equal to the sum of its previous 2 numbers.

In this article, we will learn how to identify if a given number belongs to the Fibonacci series or not.

**Examples :

**Input: 8
**Output: Yes

**Input: 31
**Output: No

Fibonacci Number Check Using a Mathematical Property

_A number **n _is a Fibonacci number if and only if **one _or **both _of ( **5*n² + 4) or (5*n² – 4) _is a **perfect square .

The above mathematical expression is derived from the closed-form expression of Fibonacci numbers (Binet’s Formula) and some number theory. It’s fast and doesn’t require generating the Fibonacci sequence. Let's look at the code implementation in Python:

Python `

import math

def is_perfect_sq(x): s = int(math.sqrt(x)) return s * s == x

def is_fibonacci(n): return is_perfect_sq(5 * n * n + 4) or is_perfect_sq(5 * n * n - 4)

for i in range(1, 7): if is_fibonacci(i): print(f"{i} is a Fibonacci Number") else: print(f"{i} is not a Fibonacci Number")

`

Output

1 is a Fibonacci Number 2 is a Fibonacci Number 3 is a Fibonacci Number 4 is not a Fibonacci Number 5 is a Fibonacci Number 6 is not a Fibonacci Number

**Explanation:

**1. is_perfect_sq(x):

• Calculates the integer **square root of **x.
• Returns **True if **x is a **perfect square, else False.

**2. is_fibonacci(n):

• Applies the mathematical identity:
• A number n is Fibonacci if **5*n² + 4 or 5*n² – 4 is a perfect square.
• Calls is**_perfect_sq() on both expressions to check this.

**3. Loop: Iterates through numbers 1 to 6 and prints whether each number is a Fibonacci number based on the result from **is_fibonacci().

Please refer this complete article on How to check if a given number is Fibonacci number? for more details!