How to Check if a Given Number is Fibonacci number - Python

 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:

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().
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