Question

C++ Programming

Use the following function to create a program that determines whether (n ^ 2 + 1) is prime for each of the primes not exceeding 1000.

bool check_prime(int input)
{
for(int i = 2; i <= sqrt(input); ++i)
{
   if(input % i == 0)
   {
       return false;
   }
}
return true;
}

Answer 1

#include <iostream>
#include <cmath>
using namespace std ;



//Use the following function to create a program that determines whether (n ^ 2 + 1) is prime
// for each of the primes not exceeding 1000.

bool check_prime(int input)
{
    for(int i = 2; i <= sqrt(input); ++i)
    {
        if(input % i == 0)
        {
            return false;
        }
    }
    return true;
}



int main() {
int input = 0 ;
    cout<< "Prime numbers less than 1000 and of the form (2^n) + 1 are as follows :  "<< endl ;



    // to generate all integers less than 1000 of the form 2^n + 1
    for (int i = 1; i < 10  /* because 2^10  = 1024  greater than 1000 */ ; ++i) {
        int number =  (int) pow(2 , i) + 1  ;

      bool result =   check_prime(number);
        if( result == true ){
            cout<< pow(2 , i) + 1 << "\t";

        }


    }


    return 0;
}