Question

Write a C++ function that inputs a positive integer and determines whether it is prime using trial division and find as many primes of the form n²+1 , where n is a positive integer, as you can.

c++/discrete structure

Answer 1

The basic idea of trial division method is to check if a number, num, has factors. This can be checked by dividing the number by every other number (excluding 1 and the number itself).

Code to copy:

#include <iostream>

using namespace std;

int main()

{

// Declare an int variable for the user input

int num;

// Initialize a flag variable to zero

int flag = 0;

// Prompt user to enter any positive integer

cout<<"Enter a positive integer greater than 1:\n";

// Read the user input

cin>>num;

// The loop will check if the number entered

// has a factor.

// Loop i from 2 to num-1

for(int i = 2; i < num; i++)

{

// Check if num is divisible by i

if(num % i == 0)

{

// Assign flag equal to 1 if a factor is found

flag = 1;

// end loop as soon as flag is equal to 1

break;

}

}

// Check if flag is equal to zero

if(flag == 0)

{

// Display the number is prime

cout<<"\nThe number, "<<num<<", is a prime number"<<endl;

}

else

{

// Display the number is not

cout<<"\nThe number, "<<num<<", is not a prime number"<<endl;

}

// As it is not possible to run a loop infinitely and no upper limit is mentioned in the question, the loop will check till the number enter by the user at the beginning of the program

int x;

bool f = 0;

for(int n=1; n<num; n++)

{

// Assign the number of the form (n*n)+1 to x

x = (n*n)+1;

// Check if the number x is a prime number

for(int m=2; m<x; m++)

{

// Check if num is divisible by i

if(x % m == 0)

{

// Assign flag equal to 1 if a factor is found

f = 1;

// end loop as soon as flag is equal to 1

break;

}

}

// Check if flag is equal to zero

if(f == 0)

{

// Display the numbers of that form

cout<<x<<" ";

}

}

return 0;

}

Screenshot of the code:

#include <iostream> using namespace std; int main() // Declare an int variable for the user input int num; // Initialize a flag variable to zero int flag = 0; // Prompt user to enter any positive integer cout<<"Enter a positive integer greater than 1: \n"; // Read the user input cin>>num; // The loop will check if the number entered // has a factor. // Loop i from 2 to num-1 for(int i = 2; i < num; i++) // Check if num is divisible by i if(num % i == 0) // Assign flag equal to 1 if a factor is found flag - 1; // end loop as soon as flag is equal to 1 break; // Check if flag is equal to zero if(flag -- ) // Display the number is prime cout<<"\nThe number, "<<num<<", is a prime number"<<endl; // Display the number is not cout<<"\nThe number, "<<num<<", is not a prime number"<<endl;

// As it is not possible to run a loop infinitely and no upper limit is mentioned in the question, the loop will check till the number enter by the user at the begining of the program int x; bool f = 0; for(int n=1; n<num; n++) // Assign the number of the form (n*n)+1 to x x = (n*n)+1; // Check if the number x is a prime number for(int m=2; m<x; m++) // Check if num is divisible by i if(x % m == 0) f = 1; // end loop as soon as flag is equal to 1 break; // Check if flag is equal to zero if(f == 0) // Display the numbers of that form cout<<x<<" "; return ;

Sample Output:

Enter a positive integer greater than 1: 29 The number, 29, is a prime number 2 5