How to check if a given number is Fibonacci number?

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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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*n² + 4) or (5*n² – 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 

PHP

<?php 
// PHP program to check if 
// x is a perfect square 
  
// A utility function that 
// returns true if x is  
// perfect square 
function isPerfectSquare($x) 
{ 
    $s = (int)(sqrt($x)); 
    return ($s * $s == $x); 
} 
  
// Returns true if n is a 
// Fibinacci Number, else false 
function 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) ||  
           isPerfectSquare(5 * $n * $n - 4); 
} 
  
// Driver Code 
for ($i = 1; $i <= 10; $i++) 
if(isFibonacci($i)) 
echo "$i is a Fibonacci Number \n"; 
else
echo "$i is a not Fibonacci Number \n" ; 
  
// This code is contributed by mits 
?> 

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

My Personal Notes

Suggest a Topic

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python

C#

PHP

Recommended Posts: