# Proof of the infinitude of primes by probabilistic methods.

I'm looking if there is proof of the infinitude of prime numbers using probabilistic method. I am motivated by the answer of my question here. The answer is based on a relationship between independence of measurable sets and integers coprime.

QUESTION: There is a proof of the infinitude of primes using the Lovász Local lemma by any of several different versions of the lemma?

-
Why would one want to use LLL here? –  Did Jul 3 '13 at 19:08
I put a "comment" after the answer by nullUser calling attention to this question. –  Will Jagy Jul 3 '13 at 19:15
Note that, as you both are anonymous, one good choice would be a chat room for the two of you. That would work pretty well if you are in similar time zones. There are instructions on the longest running chat room about how to enable "ChatJax" in order to use Latex. –  Will Jagy Jul 3 '13 at 19:18
@Did The beauty and richness of the proof of this theorem with the LLL would not be a sufficient reason? –  Elias Jul 3 '13 at 19:27
This seems to be the stuff some rather nice, innovative papers are made of: somebody wants to prove something using methods and stuff that bear no straightforward, clear conection to that something...Of course, it could perfectly well be that nobody can (at least so far) see any possible conection between things and nothing happens. –  DonAntonio Jul 3 '13 at 19:53

I don't know how to do it using your proposed lemmas, but if you would like a probabilistic proof, we can work it out from my previous answer. Again take $P(X=n) = n^{-s}/\zeta(s)$ and $E_k := \{X \text{ is divisible by } k\}$. We already showed for $s>1$ $$\left(\sum_{n=1}^\infty n^{-s}\right)^{-1}=\frac{1}{\zeta(s)} = P(X=1) = P(\cap_{p} E_p^c) = \prod_p(1-P(E_p)) = \prod_p(1-p^{-s}).$$

Assume for contradiction that there are finitely many primes. Now let $s\to 1^+$. Then we get $$0 = \prod_p(1-p^{-1})$$ which cannot be, as the right side is a finite product of strictly positive terms. Thus it must be that there are infinitely many primes.

-
I believe that the proof would be the type to show that for any natural number $N$ the likelihood of a prime greater than $N$ would be positive. –  Elias Jul 3 '13 at 19:31
@Elias That kind of argument won't work, because primes aren't random. Either there is a prime or there isn't a prime greater than $n$. Look at the proof to the previous question, and you'll see it defines a very specific random variable $X$. –  Thomas Andrews Jul 3 '13 at 19:36