According to Euclid Euler Theorem, a perfect number which is even, can be represented in the form 
where n is a prime number and 
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, it 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, it 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, it 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 = 
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 true;}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 Theoremclass 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 |
Python3
# Python3 code to verify Euclid Euler Theorem#define show(x) cout << #x << " = " << x << "\n";def isprime(n): i = 2 # check whether a number is prime or not while(i * i <= n): if (n % i == 0): return False; i += 1 return False;def isperfect(n): # perfect numbers # check is n is perfect sum of divisors # except the number itself = number s = -n; i =1 while(i * i <= n): # 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; i += 1 return (n == s);# Driver codeif __name__=='__main__': # storing powers of 2 to access in O(1) time power2 = [1<<i for i in range(61)] # generation of first few numbers # satisfying Euclid Euler's theorem print("Generating first few numbers satisfying Euclid Euler's theorem"); for i in range(2, 26): no = (power2[i] - 1) * (power2[i - 1]); if (isperfect(no) and (no % 2 == 0)): print("(2^{} - 1) * (2^({} - 1)) = {}".format(i, i, no)) # This code is contributed by rutvik_56. |
C#
// C# code to verify Euclid Euler Theoremusing 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<< $i;// generation of first few // numbers satisfying Euclid // Euler's theoremecho "Generating first few numbers " . "satisfying Euclid Euler's theorem\n"; for ($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?> |
Javascript
<script>// JavaScript program to verify Euclid Euler Theorem function isprime(n) { // check whether a number is prime or not for (let 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 let s = -n; for (let i = 1; i * i <= n; i++) { // is i is a divisor of n if (n % i == 0) { let 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 let power2 = []; for (let i = 0; i <= 60; i++) { power2[i] = 1 << i; } // generation of first few numbers // satisfying Euclid Euler's theorem document.write("Generating first few numbers " + "satisfying Euclid Euler's theorem" + "<br/>"); for (let i = 2; i <= 25; i++) { let no = (power2[i] - 1) * (power2[i - 1]); if (isperfect(no) && (no % 2 == 0)) { document.write("(2^" + i + " - 1) * (2^(" + i + " - 1)) = " + no + "<br/>"); } } // This code is contributed by code_hunt.</script> |
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
Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)
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