Check if a Given Number is Fibonacci number? (original) (raw)

Last Updated : 23 May, 2026

Given a number n, check if n is a Fibonacci number. First few Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ..

**Examples :

**Input: n = 34
**Output: true
**Explanation: 34 is one of the numbers of the Fibonacci series.

**Input: n = 41
**Output: false
**Explanation: 41 is not in the numbers of the Fibonacci series.

Try It Yourself

Table of Content

• Generate Fibonacci Numbers Iteratively - O(log n) Time O(1) Space
• Using Perfect Square Property - O(1) Time O(1) Space

Generate Fibonacci Numbers Iteratively - O(log n) Time O(1) Space

The idea is to generate Fibonacci numbers one by one starting from 0 and 1 until the generated number becomes greater than or equal to n.

• If the generated Fibonacci number becomes equal to n, then n is a Fibonacci number.
• Otherwise, n is not a Fibonacci number.

C++ `

#include <bits/stdc++.h> using namespace std;

bool isFibonacci(int n) {

// Base cases
if (n == 0 || n == 1)
{
    return true;
}

int a = 0;
int b = 1;
int c = a + b;

// Generate Fibonacci numbers
while (c < n)
{
    a = b;
    b = c;
    c = a + b;
}

return c == n;

}

// Driver Code int main() { int n = 34;

if (isFibonacci(n))
{
    cout << "true";
}
else
{
    cout << "false";
}

return 0;

}

Java

public class GfG {

public static boolean isFibonacci(int n)
{

    // Base cases
    if (n == 0 || n == 1)
    {
        return true;
    }

    int a = 0;
    int b = 1;
    int c = a + b;

    // Generate Fibonacci numbers
    while (c < n)
    {
        a = b;
        b = c;
        c = a + b;
    }

    return c == n;
}

// Driver Code
public static void main(String[] args)
{
    int n = 34;

    if (isFibonacci(n))
    {
        System.out.println("true");
    }
    else
    {
        System.out.println("false");
    }
}

}

Python

def isFibonacci(n):

# Base cases
if n == 0 or n == 1:
    return True

a = 0
b = 1
c = a + b

# Generate Fibonacci numbers
while c < n:
    a = b
    b = c
    c = a + b

return c == n

Driver Code

if name == 'main': n = 34

if isFibonacci(n):
    print('true')
else:
    print('false')

C#

using System;

public class GfG { public static bool IsFibonacci(int n) {

    // Base cases
    if (n == 0 || n == 1)
    {
        return true;
    }

    int a = 0;
    int b = 1;
    int c = a + b;

    // Generate Fibonacci numbers
    while (c < n)
    {
        a = b;
        b = c;
        c = a + b;
    }

    return c == n;
}

// Driver Code
public static void Main()
{
    int n = 34;

    if (IsFibonacci(n))
    {
        Console.WriteLine("true");
    }
    else
    {
        Console.WriteLine("false");
    }
}

}

JavaScript

function isFibonacci(n) {

// Base cases
if (n == 0 || n == 1) {
    return true;
}

let a = 0;
let b = 1;
let c = a + b;

// Generate Fibonacci numbers
while (c < n) {
    a = b;
    b = c;
    c = a + b;
}

return c == n;

}

// Driver Code let n = 34;

if (isFibonacci(n)) { console.log('true'); } else { console.log('false'); }

`

**Time Complexity: O(log n)
**Space Complexity: O(1)

Using Perfect Square Property - O(1) Time O(1) Space

A number n is a Fibonacci number if either of the following is a perfect square: 5n 2 **+ 4 or 5n 2 − 4. So, we calculate both values and check if any one of them is a perfect square. If any one of them is a perfect square, then n is a Fibonacci number.

This is a mathematical property of Fibonacci numbers: _**5n_2 **+ 4 _or _5n_2 − 4.

C++ `

#include <bits/stdc++.h> using namespace std;

// Check if x is a perfect square bool isPerfectSquare(int x) { int s = sqrt(x);

return s * s == x;

}

// Check if n is a Fibonacci number bool isFibonacci(int n) { int val1 = 5 * n * n + 4; int val2 = 5 * n * n - 4;

return isPerfectSquare(val1) || isPerfectSquare(val2);

}

int main() { int n = 34;

if (isFibonacci(n))
    cout << "true";
else
    cout << "false";

return 0;

}

Java

import java.lang.Math;

// Check if x is a perfect square public class GfG { public static boolean isPerfectSquare(int x) { int s = (int) Math.sqrt(x);

    return s * s == x;
}

// Check if n is a Fibonacci number
public static boolean isFibonacci(int n)
{
    int val1 = 5 * n * n + 4;
    int val2 = 5 * n * n - 4;

    return isPerfectSquare(val1) || isPerfectSquare(val2);
}

public static void main(String[] args)
{
    int n = 34;

    if (isFibonacci(n))
        System.out.println("true");
    else
        System.out.println("false");
}

}

Python

import math

Check if x is a perfect square

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

return s * s == x

Check if n is a Fibonacci number

def isFibonacci(n): val1 = 5 * n * n + 4 val2 = 5 * n * n - 4 return isPerfectSquare(val1) or isPerfectSquare(val2)

n = 34 if isFibonacci(n): print("true") else: print("false")

C#

using System;

// Check if x is a perfect square public class GfG { public static bool isPerfectSquare(int x) { int s = (int)Math.Sqrt(x);

    return s * s == x;
}

// Check if n is a Fibonacci number
public static bool isFibonacci(int n)
{
    int val1 = 5 * n * n + 4;
    int val2 = 5 * n * n - 4;

    return isPerfectSquare(val1) || isPerfectSquare(val2);
}

public static void Main()
{
    int n = 34;

    if (isFibonacci(n))
        Console.WriteLine("true");
    else
        Console.WriteLine("false");
}

}

JavaScript

// Check if x is a perfect square function isPerfectSquare(x) { let s = Math.sqrt(x);

return s * s === x;

}

// Check if n is a Fibonacci number function isFibonacci(n) { let val1 = 5 * n * n + 4; let val2 = 5 * n * n - 4;

return isPerfectSquare(val1) || isPerfectSquare(val2);

}

let n = 34;

if (isFibonacci(n)) { console.log('true'); } else { console.log('false'); }

`

**Time Complexity: O(1)
**Space Complexity: O(1)