If both are prime, return Yes; otherwise, return No.

Approach:

We will create two functions:
- One to check if a number is prime.
- One to check whether both n and n - 2 are prime.
In the prime-checking function, we will loop from i = 2 to sqrt(n).
If n % i == 0, then n is divisible by i and not a prime number — return false.
We will repeat the same check for n - 2.

Code

#include <bits/stdc++.h>
using namespace std;

class SumOf2Prime {
public:
    // Function to check if a number is prime
    bool isPrime(int n) {
        if (n <= 1) return false;

        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) return false;
        }

        return true;
    }

    // Function to check if n and (n - 2) are both prime
    bool prime(int n) {
        return isPrime(n) && isPrime(n - 2);
    }
};

int main() {
    int n = 19;
    SumOf2Prime s1;

    if (s1.prime(n)) {
        cout << "Yes\n";
    } else {
        cout << "No\n";
    }

    return 0;
}