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

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marked as duplicate by Steven Stadnicki, draks ..., egreg, Daryl, words that end in GRY Jun 14 '14 at 0:39

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I googled "infinitely many primes" and found this – user88595 Jun 13 '14 at 22:04
$$\pi(x) \sim \operatorname{Li} x \sim \frac{x}{\log x}$$ – Daniel Fischer Jun 13 '14 at 22:07
@BalarkaSen Is that a product over primes? How did the summation turn into that? – user45572 Jun 13 '14 at 22:08
$$\sum_{n=1}^\infty \frac1n = \prod_p \left ( 1 - \frac1p \right )^{-1}$$ The harmonic series diverges. $$\sum_{n=1}^\infty \frac1{n^2} = \pi^2/6 = \prod_p \left ( 1 - \frac1{p^2} \right)^{-1}$$ $\pi^2$ is irrational. Plus, if you forgive me the cheek $$\sum_{n=1}^\infty \frac{1}{n^3} = \prod_p \left ( 1 - \frac1{p^3}\right)^{-1}$$ $\zeta(3)$ is irrational too. – Balarka Sen Jun 13 '14 at 22:09
@DanielFischer Li? – user45572 Jun 13 '14 at 22:09
up vote 5 down vote accepted

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

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

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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.

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I think $2^{2^{n}}-1$ has the same property ( ie, at least $n$ different prime factors). – Geoff Robinson Jun 13 '14 at 23:02

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.

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(+1)This is one fun proof! Interestingly, irrationality of $\pi$ gives much stronger result than infinitude of primes $$\frac{\pi}{4} = \sum_{n= 0}^\infty \frac{(-1)^n}{2n+1} = \prod_{p \equiv 1(4)} \frac{p}{p-1} \cdot \prod_{p \equiv 3(4)} \frac{p}{p+1}$$ Thus at least one of the class $1\pmod{4}$ or $3\pmod{4}$ contains infinitely many primes. – Balarka Sen Jun 17 '14 at 9:52

There is a topological proof:

Apart from that there are almost infinitely many proofs..

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Could you explain the topological proof? I find the Wiki article difficult to parse... – user45572 Jun 13 '14 at 22:06
@user45572 Explanation: Declare each arithmetic sequence to be "closed", and extend this terminology to finite unions. Verify that the complement of a finite set is not closed. Verify that the complement of $\{\pm 1\}$ is the union of arithmetic sequences whose ratios are exactly the prime numbers (by prime factorization). Since it's cofinite, this must be an infinite union, QED. – Ryan Reich Jun 13 '14 at 22:18
Note that it's not really a topological proof at all. – Ryan Reich Jun 13 '14 at 22:23

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.

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