Round #1019

What is the most elegant way to prove that there are infinitely many prime numbers?

```csharp using System; class Program { static void Main() { // Assume there are finitely many prime numbers // p_1, p_2, ..., p_n // Multiply all prime numbers and add 1 // N = p_1 * p_2 * ... * p_n + 1 // N is either prime or composite // If N is prime, then it is a new prime number not in the list // If N is composite, then it must have a prime factor // But none of the prime numbers p_1, p_2, ..., p_n can divide N // Therefore, there exists a prime number not in the list Console.WriteLine("There are infinitely many prime numbers."); } } ```

The most elegant way to prove that there are infinitely many prime numbers is through Euclid's proof, which can be summarized as follows: assume there are finitely many primes, then multiply all of them and add 1, the resulting number is either prime or composite, if it's prime then it's a new prime not in the list, if it's composite then it must have a prime factor not in the list, either way this contradicts the assumption that the list is complete. Here is a C# implementation that demonstrates this proof: ```csharp using System; using System.Collections.Generic; public class EuclidProof { public static void Main(string[] args) { // Assume a finite list of primes List<int> primes = new List<int> { 2, 3, 5, 7, 11 }; // Multiply all primes and add 1 int n = 1; foreach (int prime in primes) { n *= prime; } n += 1; // Check if n is prime or composite if (IsPrime(n)) { Console.WriteLine("n is prime, which means it's a new prime not in the list"); } else { Console.WriteLine("n is composite, which means it has a prime factor not in the list"); } } public static bool IsPrime(int number) { if (number <= 1) return false; if (number == 2) return true; if (number % 2 == 0) return false; var boundary = (int)Math.Floor(Math.Sqrt(number)); for (int i = 3; i <= boundary; i += 2) { if (number % i == 0) return false; } return true; } } ``` This implementation demonstrates the proof by assuming a finite list of primes, multiplying them and adding 1, and then checking if the resulting number is prime or composite. If it's prime, then it's a new prime not in the list, and if it's composite, then it must have a prime factor not in the list, which contradicts the assumption that the list is complete. This proof requires more text to fully explain the underlying mathematics and the implementation in C#.