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Hi everyone.

I'd like to refer you to two papers by Arian Berdellima on odd perfect numbers:

More Properties About Odd Perfect Numbers

http://mpra.ub.uni-muenchen.de/31587/1/MPRA_paper_31587.pdf

Perfect numbers - a lower bound for an odd perfect number

http://mpra.ub.uni-muenchen.de/31218/1/MPRA_paper_31218.pdf

I have perused both papers, and the first paper in particular appears to have shown that the Euler prime of an odd perfect number must be at least 13.

Can anybody comment as to the merit(s) of these papers?

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  • $\begingroup$ In particular, in Proposition 3.2 from the first paper, am I right in thinking that $\beta \ge 0$ instead of $\beta > 0$? $\endgroup$ – Arnie Bebita-Dris Jan 28 '12 at 12:23
  • $\begingroup$ There is a meta thread on "Discussing preprints on MO" here: tea.mathoverflow.net/discussion/1059 $\endgroup$ – Gerald Edgar Jan 28 '12 at 14:08
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    $\begingroup$ The rule of thumb is that if a manuscript is typed in MS Word, then it's probably rubbish. ;) $\endgroup$ – Nikita Sidorov Jan 28 '12 at 16:52
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    $\begingroup$ This seems inappropriate, voting to close. $\endgroup$ – Igor Rivin Jan 28 '12 at 20:08
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    $\begingroup$ Arnie, I suspect proposition 3.1 does not hold, as q_j could be 1 mod q_i. Further, I do not see that 2beta + 1 has to divide q_i -1, as it could be a multiple of some factor of q_i -1. So yes, beta could be 0, and the conclusion does not follow in my view. I also do not see how the proof can be rescued. Gerhard "Ask Me About System Design" Paseman, 2012.02.01 $\endgroup$ – Gerhard Paseman Feb 1 '12 at 19:09
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Dear Arnie,

In the first paper, "More properties about the odd perfect numbers", there is a flow in the statement of Little Fermat Theorem. The modular congruences that I have used should also satisfy p=1 (mod q) and q(i)=1(mod q(j)) in order for the results to be applicable. In the second paper there is no error as it was edited by few professors and also it was the topic of my senior thesis. My apologies for the misleading result.

Regards, Arian Berdellima

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  • $\begingroup$ In the proof to your proposition 3.1, you start with "Since sigma(p^alpha) <> 2m^2 and sigma(m^2) <> p^alpha ..." . Could you expand on that part, as it is unclear to me how you get that from your assumptions in the statement of the proposition. Gerhard "Also Studies Odd Perfect Numbers" Paseman, 2012.02.01 $\endgroup$ – Gerhard Paseman Feb 1 '12 at 17:34
  • $\begingroup$ In proposition 3.1 I use the result from Pomerance that if n=p^alpha*m^2 is an odd perfect number then gcd(sigma(m^2), sigma(p^alpha))<>1,therefore for some prime q(i) which divides m^2 we have that q(i)|sigma(m^2) and q(i)|sigma(p^alpha). So, (1) 1+q(j)+...+q(j)^(2beta)=0 (mod q(i)) Notice that q(i) and q(j) are both prime factors of m^2. From (1) one obtains (2) q(j)^(2beta+1)=1 (mod q(i)) Now if q(j)=1 (mod q(i)) then there is nothing to be found because q(j)^n=1 (mod q(i)), n-positive integer. The other case is when q(j)<>1 (mod q(i)). q(j) cant be a primitive root and the results follow $\endgroup$ – Arian Feb 1 '12 at 20:11
  • $\begingroup$ In TeX: ...if $n=p^{\alpha}m^2$ is an odd perfect number then $\gcd(\sigma(m^2),\sigma(p^{\alpha}))\ne1$, therefore for some prime $q_i$ which divides $m^2$ we have that $q_i\mid\sigma(m^2)$ and $q_i\mid\sigma(p^{\alpha})$. So, (1) $1+q_j+\cdots+q_j^{2\beta}\equiv0\pmod{q_i}$. Notice that $q_i$ and $q_j$ are both prime factors of $m^2$. From (1) one obtains (2) $q_j^{2\beta+1}\equiv1\pmod{q_i}$. Now if $q_j\equiv1\pmod{q_i}$ then there is nothing to be found because $q_j^n\equiv1\pmod{q_i}$, $n$ a positive integer. The other case is when $q_j\not\equiv1\pmod{q_i}$. Continued... $\endgroup$ – Gerry Myerson Feb 1 '12 at 22:52
  • $\begingroup$ (Continuation) $q_j$ can't be a primitive root and the results follow. $\endgroup$ – Gerry Myerson Feb 1 '12 at 22:53
  • $\begingroup$ Thank you Garry for your assistance. I hope the paper makes more sense now. $\endgroup$ – Arian Feb 2 '12 at 2:23

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