Alternative Proof of Infinitely Many Primes?

Question

I've seen Euclid's proof of infinitely many primes, what are other approaches to proving there are infinitely many primes?

Answer 1

$$\prod_k \frac 1 {1-p_k^{-2}}=\sum_n n^{-2}=\zeta(2)=\frac {\pi^2}6,$$ and $\pi^2$ is irrational (from my favorite math resource) so we need infinitely many primes...

Answer 2

There is a topological proof: http://en.wikipedia.org/wiki/Furstenberg%27s_proof_of_the_infinitude_of_primes

Apart from that there are almost infinitely many proofs..

Answer 3

I like Dirichlet's theorem on primes in arithmetic progressions. Shows not only infinitely many primes, but infinitely many primes in each (nontrivial) residue class in any modulus. Proof relies on analytic properties of Dirichlet series.

Answer 4

The prime harmonic series $\frac 1 2 + \frac 1 3 + \frac 1 5 + \ldots$ diverges.

Answer 5

Here is one of my favorites: $f(n) = 2^{2^{n}}+2^{2^{n-1}}+1, n \in \mathbb{N}$ has $n$ different prime factors. Prove by induction.

Answer 6

Here's another proof.I thought of it a while ago, and posted it as an answer to a mathoverflow question, but I don't imagine I was the first to think of it. The idea is not to rely on the fact that $\pi$ is irrational. Let $\mathcal{P}$ denote the set of primes. We have $\sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$ and $\sum_{n=1}^{\infty} \frac{1}{n^{4}} = \frac{\pi^{4}}{90}.$ The first equation yields, $\prod_{p \in \mathcal{P}}\frac{p^{2}}{p^{2}-1} = \frac{\pi^{2}}{6}$ and the second leads to $\prod_{p \in \mathcal{P}}\frac{p^{4}}{p^{4}-1} = \frac{\pi^{4}}{90}.$ If we square the first of the last two equations and divide by the second, we obtain $\prod_{p \in \mathcal{P}}\frac{p^{2}+1}{p^{2}-1} = \frac{5}{2}.$ If $\mathcal{P}$ was finite, the left hand expression would be a rational number whose numerator was not divisible by $3$, but whose denominator was divisible by $3$ (because $p^{2}+1 \equiv 2$ (mod $3$) for every prime other than $3$, and $p^{2}-1$ is divisible by $3$ for every prime other than $3$). However, no such rational number could be equal to $\frac{5}{2},$ a contradiction.