c++ - Trying to print first 5 perfect numbers using Euler's theorem -

euler theorem states equation results perfect number if right side (2^n -1) prime number.

2^(n-1) * (2^n -1)

my code trying generate first 5 perfect numbers being : 6,28,496.8128 etc... upon execution prints numbers of euler's equation different n values (1,2,3,4 etc..). seems there mistake prime condition. can ?

so 6,28,120,496,2016 etc... result

#include <iostream> #include <cmath> #include <iomanip>  using namespace std; int main(void) {     (int = 2; i<100; i++)     {     bool prime = true;     (int j = 2; j*j <= i; j++)     {         if (pow(2,i) - 1 % j == 0)         {             prime = false;             }     }     if (prime = true)     {         cout << pow(2, - 1)*(pow(2, i) - 1) << " ";     } } return 0; } 

in if condition have

if (prime = true) 

the = operator assignment operator. evaluates true on every iteration in case. should using == operator. instead should have

if (prime == true) 

also, inner for loop should run until sqrt of (pow(2, i) - 1). right test if numbers i prime. instead, make it

for (int j = 2; j*j <= pow(2,1)-1; j++) 

now, iterate on numbers until square root of desired value (the "right side" of euler's theorem).

sorry edits , additions, also: in first if statement, think expression have not produce desired results, because of order of operations. in c++, % operand has higher precedence +/-, in expression

pow(2,i) - 1 % j == 0 

1 % j evaluated first, subtraction carried out. again, not produce desired results. natural solution place parentheses around expression, so:

(pow(2,i) - 1) % j == 0 

but think might create sort of "invalid operands" error, since left side of type double or float, , right int. safe, cast int:

(int)(pow(2,i) - 1) % j == 0