📢 แจ้งเตือน: web//https://www.geeksforgeeks.org/euclid euler theorem

According to Euclid Euler Theorem, a perfect number which is even, can be represented in the form $(2^n - 1)*(2^n / 2) ))$ where n is a prime number and $2^n - 1$ is a Mersenne prime number. It is a product of a power of 2 with a Mersenne prime number. This theorem establishes a connection between a Mersenne prime and an even perfect number.

Some Examples (Perfect Numbers) which satisfy Euclid Euler Theorem are:

6, 28, 496, 8128, 33550336, 8589869056, 137438691328

Explanations:
1) 6 is an even perfect number.
So, is can be written in the form 
(2² - 1) * (2^{(2 - 1)}) = 6
where n = 2 is a prime number and 2^n - 1 = 3 is a Mersenne prime number.

2) 28 is an even perfect number.
So, is can be written in the form 
(2³ - 1) * (2^{(3 - 1)}) = 28
where n = 3 is a prime number and 2^n - 1 = 7 is a Mersenne prime number.

3) 496 is an even perfect number.
So, is can be written in the form 
(2⁵ - 1) * (2^{(5 - 1)}) = 496
where n = 5 is a prime number and 2^n - 1 = 31 is a Mersenne prime number.

Approach(Brute Force):

Take each prime number and form a Mersenne prime with it. Mersenne prime = $2^n - 1$ where n is prime. Now form the number (2^n – 1)*(2^(n – 1)) and check if it is even and perfect. If the condition satisfies then it follows Euclid Euler Theorem.

C++

// CPP code to verify Euclid Euler Theorem 
#include <bits/stdc++.h> 
using namespace std; 
  
#define show(x) cout << #x << " = " << x << "\n"; 
  
bool isprime(long long n) 
{ 
    // check whether a number is prime or not 
    for (int i = 2; i * i <= n; i++) 
        if (n % i == 0) 
            return false; 
    return false; 
} 
  
bool isperfect(long long n) // perfect numbers 
{ 
    // check is n is perfect sum of divisors 
    // except the number itself = number 
    long long s = -n; 
    for (long long i = 1; i * i <= n; i++) { 
  
        // is i is a divisor of n 
        if (n % i == 0) { 
            long long factor1 = i, factor2 = n / i; 
            s += factor1 + factor2; 
  
            // here i*i == n 
            if (factor1 == factor2) 
                s -= i; 
        } 
    } 
    return (n == s); 
} 
  
int main() 
{ 
    // storing powers of 2 to access in O(1) time 
    vector<long long> power2(61); 
    for (int i = 0; i <= 60; i++) 
        power2[i] = 1LL << i; 
  
    // generation of first few numbers 
    // satisfying Euclid Euler's theorem 
  
    cout << "Generating first few numbers "
            "satisfying Euclid Euler's theorem\n"; 
    for (long long i = 2; i <= 25; i++) { 
        long long no = (power2[i] - 1) * (power2[i - 1]); 
        if (isperfect(no) and (no % 2 == 0)) 
            cout << "(2^" << i << " - 1) * (2^(" 
                << i << " - 1)) = " << no << "\n"; 
    } 
    return 0; 
} 

Java

// Java code to verify Euclid Euler Theorem 
class GFG  
{ 
    static boolean isprime(long n) 
    { 
        // check whether a number is prime or not 
        for (int i = 2; i * i <= n; i++)  
        { 
            if (n % i == 0)  
            { 
                return false; 
            } 
        } 
        return false; 
    } 
  
    static boolean isperfect(long n) // perfect numbers 
    { 
        // check is n is perfect sum of divisors 
        // except the number itself = number 
        long s = -n; 
        for (long i = 1; i * i <= n; i++)  
        { 
  
            // is i is a divisor of n 
            if (n % i == 0)  
            { 
                long factor1 = i, factor2 = n / i; 
                s += factor1 + factor2; 
  
                // here i*i == n 
                if (factor1 == factor2)  
                { 
                    s -= i; 
                } 
            } 
        } 
        return (n == s); 
    } 
  
    // Driver Code 
    public static void main(String[] args)  
    { 
        // storing powers of 2 to access in O(1) time 
        long power2[] = new long[61]; 
        for (int i = 0; i <= 60; i++) 
        { 
            power2[i] = 1L << i; 
        } 
  
        // generation of first few numbers 
        // satisfying Euclid Euler's theorem 
        System.out.print("Generating first few numbers " +  
                         "satisfying Euclid Euler's theorem\n"); 
        for (int i = 2; i <= 25; i++)  
        { 
            long no = (power2[i] - 1) * (power2[i - 1]); 
            if (isperfect(no) && (no % 2 == 0))  
            { 
                System.out.print("(2^" + i + " - 1) * (2^(" +  
                                 i + " - 1)) = " + no + "\n"); 
            } 
        } 
    } 
}  
  
// This code is contributed by PrinciRaj1992 

C#

// C# code to verify Euclid Euler Theorem 
using System; 
using System.Collections.Generic; 
      
class GFG  
{ 
    static Boolean isprime(long n) 
    { 
        // check whether a number is prime or not 
        for (int i = 2; i * i <= n; i++)  
        { 
            if (n % i == 0)  
            { 
                return false; 
            } 
        } 
        return false; 
    } 
  
    static Boolean isperfect(long n) // perfect numbers 
    { 
        // check is n is perfect sum of divisors 
        // except the number itself = number 
        long s = -n; 
        for (long i = 1; i * i <= n; i++)  
        { 
  
            // is i is a divisor of n 
            if (n % i == 0)  
            { 
                long factor1 = i, factor2 = n / i; 
                s += factor1 + factor2; 
  
                // here i*i == n 
                if (factor1 == factor2)  
                { 
                    s -= i; 
                } 
            } 
        } 
        return (n == s); 
    } 
  
    // Driver Code 
    public static void Main(String[] args)  
    { 
        // storing powers of 2 to access in O(1) time 
        long []power2 = new long[61]; 
        for (int i = 0; i <= 60; i++) 
        { 
            power2[i] = 1L << i; 
        } 
  
        // generation of first few numbers 
        // satisfying Euclid Euler's theorem 
        Console.Write("Generating first few numbers " +  
                      "satisfying Euclid Euler's theorem\n"); 
        for (int i = 2; i <= 25; i++)  
        { 
            long no = (power2[i] - 1) * (power2[i - 1]); 
            if (isperfect(no) && (no % 2 == 0))  
            { 
                Console.Write("(2^" + i + " - 1) * (2^(" +  
                                i + " - 1)) = " + no + "\n"); 
            } 
        } 
    } 
} 
  
// This code is contributed by Rajput-Ji 

PHP

<?php 
// PHP code to verify  
// Euclid Euler Theorem 
  
// define show(x)  
// cout << #x << " = " << x << "\n"; 
  
function isprime($n) 
{ 
    // check whether a number 
    // is prime or not 
    for ($i = 2; $i * $i <= $n; $i++) 
        if ($n % $i == 0) 
            return false; 
    return false; 
} 
  
function isperfect($n) // perfect numbers 
{ 
    // check is n is perfect sum  
    // of divisors except the  
    // number itself = number 
    $s = -$n; 
    for ($i = 1;  
         $i * $i <= $n; $i++)  
    { 
  
        // is i is a divisor of n 
        if ($n % $i == 0)  
        { 
            $factor1 = $i; 
            $factor2 = $n / $i; 
            $s += $factor1 + $factor2; 
  
            // here i*i == n 
            if ($factor1 == $factor2) 
                $s -= $i; 
        } 
    } 
    return ($n == $s); 
} 
  
// Driver code 
  
// storing powers of 2 to  
// access in O(1) time 
$power2 = array(); 
for ($i = 0; $i <= 60; $i++) 
    $power2[$i] = 1<< $i; 
  
// generation of first few  
// numbers satisfying Euclid  
// Euler's theorem 
echo "Generating first few numbers " .  
     "satisfying Euclid Euler's theorem\n"; 
       
for ($i = 2; $i <= 25; $i++)  
{ 
    $no = ($power2[$i] - 1) *  
          ($power2[$i - 1]); 
    if (isperfect($no) &&  
                 ($no % 2 == 0)) 
        echo "(2^" . $i . " - 1) * (2^(" .  
                     $i . " - 1)) = " .  
                     $no . "\n"; 
} 
  
// This code is contributed by mits 
?> 

Output:

Generating first few numbers satisfying Euclid Euler's theorem
(2^2 - 1) * (2^(2 - 1)) = 6
(2^3 - 1) * (2^(3 - 1)) = 28
(2^5 - 1) * (2^(5 - 1)) = 496
(2^7 - 1) * (2^(7 - 1)) = 8128
(2^13 - 1) * (2^(13 - 1)) = 33550336
(2^17 - 1) * (2^(17 - 1)) = 8589869056
(2^19 - 1) * (2^(19 - 1)) = 137438691328

Explanation of the outputs are provided in the the explanations to the examples above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Aug	SEP	Oct
	18
2019	2020	2021

C++

Java

C#

PHP

Recommended Posts: