Are there infinitely many prime numbers?

Question

I will post my own answer below. That should not deter others from answering. There are many ways to prove this.

Answer 1

The number $1+\Big(2\times17\times61\times97\Big)$ does not have $2$, $17$, $61$, or $97$ among its prime factors, because when divided by any of them, it leaves a remainder of $1$. Either it is prime, or it has prime factors not among those listed. Either way, there is at least one more prime than those listed.

With any finite set of primes, you can find at least one more prime not among them, by proceding as above.

This argument was written by Euclid of Alexandria (after whom Euclidean geometry and the Euclidean algorithm are named) in the third century BC.

Many authors rearrange it into a proof by contradiction, thus: suppose only finitely many prime numbers exist; multiply them and then add $1$, getting a number not divisible by any prime numbers. But every positive integer (except $1$) is divisible by some primes, so our assumption that only finitely many exist has led to a contradiction. However, making it a proof by contradiction merely adds an extra complication that doesn't help. Many authors who write it that way mistakenly say that Euclid wrote it as a proof by contradiction. Mathematicians are better at mathematics than at history.

In many published accounts, instead of an arbitrary set of prime numbers, one uses the smallest $n$ prime numbers, for some $n$. The conclusion is then given thus: there is always at least one prime number larger than the one used at the outset. But Euclid merely said there is always at least one other than the ones used at the outset. I now see that the number mentioned above, $1+\Big(2\times17\times61\times97\Big)$, is divisible by $11$. In some cases, some or all of the primes other than those in the initial set are smaller than all of the primes in the inital set. That happens, for example, if the initial set is $\{5,7\}$.