Python Program for How to check if a given number is a Fibonacci number?

In this article, we will learn how to check if a given number is a Fibonacci number using a mathematical property rather than generating the entire Fibonacci sequence.

Problem Statement

Given a number n, we need to determine whether n is a Fibonacci number or not.

The Fibonacci sequence starts with 0 and 1, where each subsequent number is the sum of the previous two numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

Mathematical Property

A number is Fibonacci if and only if (5*n² + 4) or (5*n² - 4) is a perfect square. This mathematical property allows us to check Fibonacci numbers efficiently without generating the sequence.

Implementation

Helper Function to Check Perfect Square

First, we create a function to check if a number is a perfect square ?

import math

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

# Test the function
print(isPerfectSquare(25))  # True
print(isPerfectSquare(26))  # False

True
False

Main Function to Check Fibonacci Number

Now we implement the main function that uses the mathematical property ?

import math

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

def isFibonacci(n):
    # Check if either (5*n² + 4) or (5*n² - 4) is a perfect square
    return isPerfectSquare(5*n*n + 4) or isPerfectSquare(5*n*n - 4)

# Test with some numbers
test_numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

for num in test_numbers:
    if isFibonacci(num):
        print(f"{num} is a Fibonacci Number")
    else:
        print(f"{num} is not a Fibonacci Number")

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
7 is not a Fibonacci Number
8 is a Fibonacci Number
9 is not a Fibonacci Number
10 is not a Fibonacci Number

How It Works

The algorithm works by:

1. Calculating 5*n² + 4 and 5*n² - 4 for the given number n
2. Checking if either of these values is a perfect square
3. If at least one is a perfect square, then n is a Fibonacci number

Testing with Larger Numbers

Let's test the function with some larger Fibonacci numbers ?

import math

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

def isFibonacci(n):
    return isPerfectSquare(5*n*n + 4) or isPerfectSquare(5*n*n - 4)

# Test with known Fibonacci numbers
fibonacci_numbers = [13, 21, 34, 55, 89]
non_fibonacci = [14, 22, 35, 56, 90]

print("Testing known Fibonacci numbers:")
for num in fibonacci_numbers:
    result = "?" if isFibonacci(num) else "?"
    print(f"{num}: {result}")

print("\nTesting non-Fibonacci numbers:")
for num in non_fibonacci:
    result = "?" if isFibonacci(num) else "?"
    print(f"{num}: {result}")

Testing known Fibonacci numbers:
13: ?
21: ?
34: ?
55: ?
89: ?

Testing non-Fibonacci numbers:
14: ?
22: ?
35: ?
56: ?
90: ?

Conclusion

This mathematical approach provides an efficient way to check if a number is Fibonacci without generating the entire sequence. The method uses the property that a number n is Fibonacci if either (5*n² + 4) or (5*n² - 4) is a perfect square.