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http://arxiv.org/abs/0908.4287 I could not find any reviews for it, but if true its a major claim, because it says that $\Re(\rho) < 21/40$ where $\rho$ is a zeta zero.

My question:

Are there any reviews of this paper, that reject or accept the claims made in this paper? Any references will be highly appreciated.

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    $\begingroup$ Doubtful. I speculate the proofs are likely to be incorrect, although I haven't read the paper and can't say that for sure. Unfortunately it is fairly common to see incorrect proofs of related results on the arXiv. If the results were correct, I imagine the author probably would have submitted it for publication and gotten it past a referee. $\endgroup$ Apr 12, 2012 at 12:42
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    $\begingroup$ This author has claimed other fantastic results on the arXiv, with basic/silly mistakes in the proof. At one instance the nature of the error went like this: $A+B<C$, hence $A<C$ and $B<C$. Needless to say, $A$ and $B$ were not positive there. Ignore this paper safely. $\endgroup$
    – GH from MO
    Apr 12, 2012 at 14:36
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    $\begingroup$ Following tea.mathoverflow.net/discussion/1422/… , we have a reasonably general consensus that MO is not an appropriate venue to discuss recent preprints in detail and that authors should be treated with respect. $\endgroup$
    – S. Carnahan
    Aug 18, 2012 at 4:12
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    $\begingroup$ Is there any way to remove the polluting long series of articles in arxiv each full of mistakes by this same author? $\endgroup$ Jul 2, 2020 at 10:34
  • $\begingroup$ I believe arxiv can take an action - at least it can, in theory, revoke the endorser status of the author (who does have this status for math.GM). $\endgroup$ Jun 11, 2021 at 9:02

2 Answers 2

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Well,

In this case it is a short elementary paper (6 pages) and it is easy to verify, although just the length of the paper and the importance of the result and some other aspects of the paper, such as the fact that it is in general mathematics part of arxiv and it does not seem as the autor uses latex is sufficient to be quite sceptical about the claim. It is easy to find mistakes. The paper has 6 pages. The first part seems to contain some standard results in the area. The proof of the new claimed very strong result, Theorem 1 is just one page long, and it starts on page 4, so it is sufficient to read that part. The author tries to use the fact that $$ \theta(x+y)-\theta(x)>0, $$ where $$ \theta(x)=\sum_{ p < x } \log(p),$$ whenever $y \gg x^{21/40} $, which is a nice result of Baker-Harman-Pintz to prove the claimed zero free region. This is theorem 1 in the paper. It is easy to see that the claimed proof gives no such result. The author gives the elementary inequality $$ \theta(x+y)-\theta(x) < y (\log x)^3.$$ This is certainly true in relevant ranges of $x$ and $y$. However the author then somehow uses this inequality in the wrong direction into something that essentially boils down to (Eq 4 in the paper does not hold in general) $$ 0< \theta(x+y)-\theta(x) < y (\log x)^3 < \theta(x+y)-\theta(x). $$ Since this is obviously false the author obtains several contradictions, one of which is supposed to prove Theorem 1 (at least that is my guess how the author comes to that conclusion. The "proof" is not clear at all).

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To MO Forum:

If this group is doing papers reviews, it should do objective reviews.

The ones posted about my paper are clearly subjective, and probably of predetermined conclusions.

As this is a very simple, and very elementary result, this intellectual, and resourceful forum should have problem refuting it, and providing real technical flaws.

Name a sequence of zeros, or a zero on the critical strip that contradict that this result. Which ?

After this forum provides a real technical flaw, I will withdraw the paper from the arxiv.

Thank you, N. A. Carella

"... Such a fundamental and far-reaching theorem proved by so simple and elementary methods—it is pure magic.", Atle Selberg, 2005.

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    $\begingroup$ Would you comment on Eq. 4 as mentioned in the post? It may be that there is a real technical flaw stemming from that. Gerhard Paseman, 2012.08.17 $\endgroup$ Aug 17, 2012 at 23:50

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