# Question Re: Arian Berdellima's Papers On Odd Perfect Numbers [closed]

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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## closed as not constructive by Igor Rivin, Felipe Voloch, Anthony Quas, Franz Lemmermeyer, Yemon ChoiFeb 1 '12 at 18:47

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In particular, in Proposition 3.2 from the first paper, am I right in thinking that $\beta \ge 0$ instead of $\beta > 0$? –  Arnie Dris Jan 28 '12 at 12:23
There is a meta thread on "Discussing preprints on MO" here: tea.mathoverflow.net/discussion/1059 –  Gerald Edgar Jan 28 '12 at 14:08
The rule of thumb is that if a manuscript is typed in MS Word, then it's probably rubbish. ;) –  Nikita Sidorov Jan 28 '12 at 16:52
This seems inappropriate, voting to close. –  Igor Rivin Jan 28 '12 at 20:08
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 –  Gerhard Paseman Feb 1 '12 at 19:09

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... –  Gerry Myerson Feb 1 '12 at 22:52
(Continuation) $q_j$ can't be a primitive root and the results follow. –  Gerry Myerson Feb 1 '12 at 22:53