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I have been working on Odd Perfect Numbers for a while now. When I started to go through recent publications, I saw that a large number of papers have been proving results using the Factor Chain argument and computation. I am unable to understand the arguments that have been stated in those papers. Please explain the basics of the concept. If possible, provide links where the concept is explained in detail but in a basic way(because I am new to this idea).

HV6
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The following MSE question contains such an example of a factor chain argument.

Basically, the factor chain argument makes use of the simple observation that $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=2m=2 q^k n^2,$$ if $m = q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Therefore, if we assume that a squared prime $p^2 \parallel n^2$ (where $p^2 \parallel n^2$ means $p^2 \mid n^2$ but $p^3 \nmid n^2$), then via factor/sigma chains, $\sigma(p^2)=p^2+p+1 \mid n^2$.

The primes $p$ obtained in this manner are called assumed factors, while the divisors of $\sigma(p^2)$ at each step of the factor/sigma chain approach are called consequent factors.

Note that necessarily we have $p \neq q$, since $\gcd(q,n)=1$.

Factor-chain based algorithms are also discussed in pages $96$ to $97$ of Sorli's thesis, available online here.

Jose Arnaldo Bebita Dris
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  • I read the answer to the other question(whose link you have posted above). I have 2 doubts: 1) At each step, how have you gone from p^2 | m^2 to p^2 || m^2? 2) When you have got 3^2 || m^2, meaning 3^2 || N, isn't getting more than two 3s on the Sigma side of the equation a contradiction? – HV6 Sep 02 '20 at 06:09
  • @HrishikeshVenkataraman, (1) You will have to assume $p^2 \parallel m^2$ at each node of the factor/sigma chain, hence the term assumed factor. (2) Your question is a valid concern, and is essentially what I was asking for in the hyperlinked question. The short answer to your question is: Yes, if $3^2$ is an assumed factor, then getting more than two $3$'s on the $\sigma$-side of the equation would be a contradiction, subject to the restriction that you keep track of which are assumed factors and which are consequent factors. – Jose Arnaldo Bebita Dris Sep 02 '20 at 08:08
  • I went through the 2 pages in the thesis paper that you had mentioned. There was a mention of DoP, but I didn't find a definition of what DoP(...) was. Could you please explain what that term refers to. – HV6 Sep 02 '20 at 14:05
  • I did try searching for the string DoP in Sorli's thesis document, but could not find anything relevant. I did see at least two (2) DoP's from pages $96$ to $97$. I suggest you contact Sorli directly to ask about this. – Jose Arnaldo Bebita Dris Sep 03 '20 at 04:01