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I know about

"Simple analytic proof of the prime number theorem" Newman, 1980

However, is there a proof of the Prime Number Theorem without the use of complex analysis? (Real analysis is fine).


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Check this out: Erdos and Selberg have an elementary proof. – Jon Bannon Apr 4 '11 at 20:54
Yes. Do a net search. Erdos and Selberg (if memory serves) each did a mostly non-analytic version. Gerhard "Ask Me About System Design" Paseman, 2011.04.04 – Gerhard Paseman Apr 4 '11 at 20:55
Indeed, Erdos and Selberg are cited in the second sentence of Newman's paper... – j.c. Apr 4 '11 at 21:17
It is worthwhile to note that the Erdős-Selberg proof is nicely explained in Hardy-Wright: An introduction to the theory of numbers. See sections 22.14-22.16 there, especially Theorems 430 and 434. – GH from MO Apr 4 '11 at 21:18

5 Answers 5

up vote 6 down vote accepted explains the classic proof in context (there is what amounts to a priority dispute).

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For Selberg's side of the story, see the interview with him in BAMS 2008#4. – Harald Hanche-Olsen Apr 5 '11 at 9:39

Another exposition of an elementary proof (that is, a proof not using complex analysis) is in Gerald Tenenbaum and Michel Mendes France, The Prime Numbers and Their Distribution, which is Volume 6 of the Student Mathematical Library, published by the American Mathematical Society. The proof they give is due to Daboussi, from 1984.

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Somehow, I thought there is essentially just one "elementary" proof known, that due to Erdős and Selberg; do you mean, the proof you mention is substantially different? – Seva Apr 5 '11 at 7:44
@Seva: From the MR review of Daboussi's paper (written by Diamond): "This paper gives an elementary proof of the PNT that is remarkable in that it makes no use of Selberg's now famous formula." Here is a link to the original paper of Daboussi (And then there are the other devlopments improving on the error term, Diamon-Steinig and so on.) – quid Apr 5 '11 at 9:30

If you just want $\pi(n) = \Omega \left( \frac{n}{\log n} \right)$, good enough for many applications, here is a quick proof: The highest power of a prime $p$ dividing $2n \choose n$ is at most $2n$ -- you get at most one more factor of $p$ in the numerator than denominator for each power $p^i \leq 2n$. This tells you that ${2n \choose n} \leq (2n)^{\pi(2n)}$. So $\pi(2n) \geq \frac{\log_2 {2n \choose n}}{\log_2 (2n)} \geq \frac{n}{\log_2 (2n)}$.

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There is a terrific exposition of the elementary proof by Terry Tao, available as the file prime.dvi here. A more traditional exposition is available in Edwards's book Riemann's zeta function.

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A nice exposition of an Erdos/Selberg-type elementary proof is given by Levinson in Amer. Math. Monthly 76 (1969) 225–245.

The proof by Daboussi as written up by Tenenbaum and Mendes-France was already mentioned.

Yet another one is due to Hildebrand in Mathematika 33 (1986) 23–30.

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