Find out the prime numbers in the form of A+nB or B+nA

Given two integers A and B and an integer N. The task is to find out N prime numbers of the form A + nB or B + nA( n=1, 2, 3…). If it is not possible, print -1.

Examples:

Input: A = 3, B = 5, N = 4
Output: 13, 11, 17, 23
Explanation:
13 (3+2*5)
11 (5+2*3)
17 (5+4*3)
23 (3+4*5)



Input: A = 4, B = 6, N = 4
Output: -1

Approach:
Since A + nB to be a prime one thing is sure that there should not present any common factor between A and B, means A and B should be co-prime.

The best and most efficient approach will be to use Dirichlet’s Theorem.

Dirichlet’s Theorem says that if a and b are relatively prime positive integers, then the arithmetic sequence a, a+b, a+2b, a+3b…contains infinitely many primes.

So firstly check if A and B are co-prime.
If A and B are co-prime then check A+nB and B+nA for their primality where n=1, 2, 3…. Print the first N prime numbers among them.
Below is the implementation of the above approach:

C++

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// C++ implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Utility function to check
// whether two numbers is
// co-prime or not
int coprime(int a, int b)
{
    if (__gcd(a, b) == 1)
        return true;
    else
        return false;
}
  
// Utility function to check
// whether a number is prime
// or not
bool isPrime(int n)
{
    // Corner case
    if (n <= 1)
        return false;
  
    if (n == 2 or n == 3)
        return true;
  
    // Check from 2 to sqrt(n)
    for (int i = 2; i * i <= n; i++)
        if (n % i == 0)
            return false;
  
    return true;
}
  
// finding the Prime numbers
void findNumbers(int a, int b, int n)
{
  
    bool possible = true;
  
    // Checking whether given
    // numbers are co-prime
    // or not
    if (!coprime(a, b))
        possible = false;
  
    int c1 = 1;
    int c2 = 1;
  
    int num1, num2;
  
    // To store the N primes
    set<int> st;
    // If 'possible' is true
    if (possible) {
  
        // Printing n numbers
        // of prime
        while ((int)st.size() != n) {
  
            // checking the form of a+nb
            num1 = a + (c1 * b);
            if (isPrime(num1)) {
                st.insert(num1);
            }
            c1++;
  
            // Checking the form of b+na
            num2 = b + (c2 * a);
            if (isPrime(num2)) {
                st.insert(num2);
            }
            c2++;
        }
  
        for (int i : st)
            cout << i << " ";
    }
  
    // If 'possible' is false
    // return -1
    else
        cout << "-1";
}
  
// Driver Code
int main()
{
  
    int a = 3;
    int b = 5;
    int n = 4;
  
    findNumbers(a, b, n);
    return 0;
}

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Java














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// Java implementation of the approach 


import java.util.*; 


class GFG 


{ 


static int __gcd(int a, int b)  


{  


    if (b == 0)  


        return a;  


    return __gcd(b, a % b);  


} 


  


// Utility function to check 


// whether two numbers is 


// co-prime or not 


static boolean coprime(int a, int b) 


{ 


    if (__gcd(a, b) == 1) 


        return true; 


    else


        return false; 


} 


  


// Utility function to check 


// whether a number is prime 


// or not 


static boolean isPrime(int n) 


{ 


    // Corner case 


    if (n <= 1) 


        return false; 


  


    if (n == 2 || n == 3) 


        return true; 


  


    // Check from 2 to sqrt(n) 


    for (int i = 2; i * i <= n; i++) 


        if (n % i == 0) 


            return false; 


  


    return true; 


} 


  


// finding the Prime numbers 


static void findNumbers(int a, int b, int n) 


{ 


    boolean possible = true; 


  


    // Checking whether given 


    // numbers are co-prime 


    // or not 


    if (!coprime(a, b)) 


        possible = false; 


  


    int c1 = 1; 


    int c2 = 1; 


  


    int num1, num2; 


  


    // To store the N primes 


    HashSet<Integer> st = new HashSet<Integer>(); 


    // If 'possible' is true 


    if (possible) 


    { 


  


        // Printing n numbers 


        // of prime 


        while ((int)st.size() != n) 


        { 


  


            // checking the form of a+nb 


            num1 = a + (c1 * b); 


            if (isPrime(num1)) 


            { 


                st.add(num1); 


            } 


            c1++; 


  


            // Checking the form of b+na 


            num2 = b + (c2 * a); 


            if (isPrime(num2))  


            { 


                st.add(num2); 


            } 


            c2++; 


        } 


  


        for (int i : st) 


            System.out.print(i + " "); 


    } 


  


    // If 'possible' is false 


    // return -1 


    else


        System.out.print("-1"); 


} 


  


// Driver Code 


public static void main(String[] args)  


{ 


    int a = 3; 


    int b = 5; 


    int n = 4; 


  


    findNumbers(a, b, n); 


} 


} 


  


// This code is contributed by Rajput-Ji 



















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# Python3 implementation of the above approach  


from math import gcd, sqrt 


  


# Utility function to check  


# whether two numbers is  


# co-prime or not  


def coprime(a, b) : 


      


    if (gcd(a, b) == 1) : 


        return True;  


          


    else : 


        return False;  


  


# Utility function to check  


# whether a number is prime  


# or not  


def isPrime(n) :  


  


    # Corner case  


    if (n <= 1) : 


        return False;  


  


    if (n == 2 or n == 3) : 


        return True;  


  


    # Check from 2 to sqrt(n)  


    for i in range(2, int(sqrt(n)) + 1) : 


        if (n % i == 0) : 


            return False;  


  


    return True;  


  


# finding the Prime numbers  


def findNumbers(a, b, n) : 


  


    possible = True;  


  


    # Checking whether given  


    # numbers are co-prime  


    # or not  


    if (not coprime(a, b)) : 


        possible = False;  


  


    c1 = 1;  


    c2 = 1;  


  


    num1 = 0; 


    num2 = 0;  


  


    # To store the N primes  


    st = set();  


      


    # If 'possible' is true  


    if (possible) : 


  


        # Printing n numbers  


        # of prime  


        while (len(st) != n) : 


  


            # checking the form of a+nb  


            num1 = a + (c1 * b);  


              


            if (isPrime(num1)): 


                  


                st.add(num1);  


                  


            c1 += 1;  


  


            # Checking the form of b+na  


            num2 = b + (c2 * a);  


              


            if (isPrime(num2)): 


                st.add(num2);  


      


            c2 += 1;  


  


        for i in st : 


            print(i, end = " ");  


  


    # If 'possible' is false  


    # return -1  


    else : 


        print("-1");  


  


# Driver Code  


if __name__ == "__main__" :  


  


    a = 3;  


    b = 5;  


    n = 4;  


  


    findNumbers(a, b, n);  


  


# This code is contributed by AnkitRai01 



















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C#














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// C# implementation of the approach 


using System; 


using System.Collections.Generic; 


  


class GFG 


{ 


static int __gcd(int a, int b)  


{  


    if (b == 0)  


        return a;  


    return __gcd(b, a % b);  


} 


  


// Utility function to check 


// whether two numbers is 


// co-prime or not 


static bool coprime(int a, int b) 


{ 


    if (__gcd(a, b) == 1) 


        return true; 


    else


        return false; 


} 


  


// Utility function to check 


// whether a number is prime 


// or not 


static bool isPrime(int n) 


{ 


    // Corner case 


    if (n <= 1) 


        return false; 


  


    if (n == 2 || n == 3) 


        return true; 


  


    // Check from 2 to sqrt(n) 


    for (int i = 2; i * i <= n; i++) 


        if (n % i == 0) 


            return false; 


  


    return true; 


} 


  


// finding the Prime numbers 


static void findNumbers(int a, int b, int n) 


{ 


    bool possible = true; 


  


    // Checking whether given 


    // numbers are co-prime 


    // or not 


    if (!coprime(a, b)) 


        possible = false; 


  


    int c1 = 1; 


    int c2 = 1; 


  


    int num1, num2; 


  


    // To store the N primes 


    HashSet<int> st = new HashSet<int>(); 


      


    // If 'possible' is true 


    if (possible) 


    { 


  


        // Printing n numbers 


        // of prime 


        while (st.Count != n) 


        { 


  


            // checking the form of a+nb 


            num1 = a + (c1 * b); 


            if (isPrime(num1)) 


            { 


                st.Add(num1); 


            } 


            c1++; 


  


            // Checking the form of b+na 


            num2 = b + (c2 * a); 


            if (isPrime(num2))  


            { 


                st.Add(num2); 


            } 


            c2++; 


        } 


  


        foreach (int i in st) 


            Console.Write(i + " "); 


    } 


  


    // If 'possible' is false 


    // return -1 


    else


        Console.Write("-1"); 


} 


  


// Driver Code 


public static void Main(String[] args)  


{ 


    int a = 3; 


    int b = 5; 


    int n = 4; 


  


    findNumbers(a, b, n); 


} 


} 


  


// This code is contributed by 29AjayKumar 



















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Output:


11 13 17 23








    

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