Given a number ‘n’, how to check if n is a Fibonacci number. First few Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, ..

Examples :

Input : 8 Output : Yes Input : 34 Output : Yes Input : 41 Output : No

A simple way is to generate Fibonacci numbers until the generated number is greater than or equal to ‘n’. Following is an interesting property about Fibonacci numbers that can also be used to check if a given number is Fibonacci or not.

*A number is Fibonacci if and only if one or both of (5*n ^{2} + 4) or (5*n^{2} – 4) is a perfect square* (Source: Wiki). Following is a simple program based on this concept.

## C++

// C++ program to check if x is a perfect square #include <iostream> #include <math.h> using namespace std; // A utility function that returns true if x is perfect square bool isPerfectSquare(int x) { int s = sqrt(x); return (s*s == x); } // Returns true if n is a Fibinacci Number, else false bool isFibonacci(int n) { // n is Fibinacci if one of 5*n*n + 4 or 5*n*n - 4 or both // is a perferct square return isPerfectSquare(5*n*n + 4) || isPerfectSquare(5*n*n - 4); } // A utility function to test above functions int main() { for (int i = 1; i <= 10; i++) isFibonacci(i)? cout << i << " is a Fibonacci Number \n": cout << i << " is a not Fibonacci Number \n" ; return 0; }

## Java

// Java program to check if x is a perfect square class GFG { // A utility method that returns true if x is perfect square static boolean isPerfectSquare(int x) { int s = (int) Math.sqrt(x); return (s*s == x); } // Returns true if n is a Fibonacci Number, else false static boolean isFibonacci(int n) { // n is Fibonacci if one of 5*n*n + 4 or 5*n*n - 4 or both // is a perfect square return isPerfectSquare(5*n*n + 4) || isPerfectSquare(5*n*n - 4); } // Driver method public static void main(String[] args) { for (int i = 1; i <= 10; i++) System.out.println(isFibonacci(i) ? i + " is a Fibonacci Number" : i + " is a not Fibonacci Number"); } } //This code is contributed by Nikita Tiwari

## Python

# python program to check if x is a perfect square import math # A utility function that returns true if x is perfect square def isPerfectSquare(x): s = int(math.sqrt(x)) return s*s == x # Returns true if n is a Fibinacci Number, else false def isFibonacci(n): # n is Fibinacci if one of 5*n*n + 4 or 5*n*n - 4 or both # is a perferct square return isPerfectSquare(5*n*n + 4) or isPerfectSquare(5*n*n - 4) # A utility function to test above functions for i in range(1,11): if (isFibonacci(i) == True): print i,"is a Fibonacci Number" else: print i,"is a not Fibonacci Number "

## C#

// C# program to check if // x is a perfect square using System; class GFG { // A utility function that returns // true if x is perfect square static bool isPerfectSquare(int x) { int s = (int)Math.Sqrt(x); return (s * s == x); } // Returns true if n is a // Fibonacci Number, else false static bool isFibonacci(int n) { // n is Fibonacci if one of // 5*n*n + 4 or 5*n*n - 4 or // both are a perfect square return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4); } // Driver method public static void Main() { for (int i = 1; i <= 10; i++) Console.WriteLine(isFibonacci(i) ? i + " is a Fibonacci Number" : i + " is a not Fibonacci Number"); } } // This code is contributed by Sam007

Output:

1 is a Fibonacci Number 2 is a Fibonacci Number 3 is a Fibonacci Number 4 is a not Fibonacci Number 5 is a Fibonacci Number 6 is a not Fibonacci Number 7 is a not Fibonacci Number 8 is a Fibonacci Number 9 is a not Fibonacci Number 10 is a not Fibonacci Number

This article is contributed by **Abhay Rathi**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above