Legendre's Conjecture (original) (raw)

Last Updated : 13 Nov, 2022

It says that there is always one prime number between any two consecutive natural number's(n = 1, 2, 3, 4, 5, ...) square. This is called Legendre's Conjecture.

Conjecture: A conjecture is a proposition or conclusion based upon incomplete information to which no proof has been found i.e it has not been proved or disproved.

Mathematically,
there is always one prime p in the range n^2 to (n + 1)^2 where n is any natural number.
for examples-
2 and 3 are the primes in the range 1^2 to 2^2 .
5 and 7 are the primes in the range 2^2 to 3^2 .
11 and 13 are the primes in the range 3^2 to 4^2 .
17 and 19 are the primes in the range 4^2 to 5^2 .

Examples:

Input : 4 output: Primes in the range 16 and 25 are: 17 19 23

Explanation: Here 42 = 16 and 52 = 25
Hence, prime numbers between 16 and 25 are 17, 19 and 23.

Input : 10 Output: Primes in the range 100 and 121 are: 101 103 107 109 113

C++ `

// C++ program to verify Legendre's Conjecture // for a given n. #include <bits/stdc++.h> using namespace std;

// prime checking bool isprime(int n) { for (int i = 2; i * i <= n; i++) if (n % i == 0) return false; return true; }

void LegendreConjecture(int n) { cout << "Primes in the range "<<nn << " and "<<(n+1)(n+1) <<" are:" <<endl;

for (int i = nn; i <= ((n+1)(n+1)); i++)

  // searching for primes
  if (isprime(i))
      cout << i <<endl;

}

// Driver program int main() { int n = 50; LegendreConjecture(n); return 0; }

Java

// Java program to verify Legendre's Conjecture // for a given n. class GFG {

// prime checking static boolean isprime(int n) { for (int i = 2; i * i <= n; i++) if (n % i == 0) return false; return true; }

static void LegendreConjecture(int n) { System.out.println("Primes in the range "+nn +" and "+(n+1)(n+1) +" are:");

 for (int i = n*n; i <= ((n+1)*(n+1)); i++)
 {
   // searching for primes
   if (isprime(i))
     System.out.println(i);
 }

}

// Driver program public static void main(String[] args) { int n = 50; LegendreConjecture(n); } } //This code is contributed by //Smitha Dinesh Semwal

Python3

Python3 program to verify Legendre's Conjecture

for a given n

import math

def isprime( n ):

i = 2
for i in range (2, int((math.sqrt(n)+1))):
    if n%i == 0:
        return False
return True

def LegendreConjecture( n ): print ( "Primes in the range ", nn , " and ", (n+1)(n+1) , " are:" )

for i in range (n*n, (((n+1)*(n+1))+1)):
    if(isprime(i)):
        print (i)
        

n = 50 LegendreConjecture(n)

Contributed by _omg

C#

// C# program to verify Legendre's // Conjecture for a given n. using System;

class GFG {

// prime checking
static Boolean isprime(int n)
{ 
    for (int i = 2; i * i <= n; i++)
        if (n % i == 0)
            return false;
            
    return true;
}

static void LegendreConjecture(int n)
{
    Console.WriteLine("Primes in the range "
       + n * n + " and " + (n + 1) * (n + 1)
                                  + " are:");
    
    for (int i = n * n; i <= ((n + 1) 
                            * (n + 1)); i++)
    {
        
        // searching for primes
        if (isprime(i))
            Console.WriteLine(i);
    }
}

// Driver program
public static void Main(String[] args)
{
    int n = 50;
    
    LegendreConjecture(n);
}

}

// This code is contributed by parashar.

PHP

JavaScript

`

Output :

Primes in the range 2500 and 2601 are: 2503 2521 2531 2539 2543 2549 2551 2557 2579 2591 2593

Time Complexity: O(n2). isPrime() function takes O(n) time and it is embedded in LegendreConjecture() function which also takes O(n) time as it has loop which starts from n2 and ends at
(n+1)2 so, (n+1)2 - n2 = 2n+1.

Auxiliary Space: O(1)