You can read a good summary of my work on odd perfect numbers from this link: http://arxiv.org/abs/1206.3230

Here are some (fresh) new results:

Let N = (p^k)*(m^2) be an OPN with Euler factor p^k. Since N is perfect, it cannot happen that p^k = m^2. In particular, it cannot be the case that p^k = m. In other words, p = m is NOT TRUE.

Thus, either p < m or m < p. We consider the second case.

m < p <= p^k < m^2 < p^2

--> p <= p^k < p^2

--> 1 <= k < 2

--> k = 1

That is, we have shown that: m < p <= p^k implies that k = 1.

The contrapositive is: k >= 5 --> p <= p^k < m. (i.e. k is congruent to 1 modulo 4)

Now, I am still trying to figure out the constraints that the first inequality would impose on the canonical factorization of N. At any rate, would it imply that the Euler prime is the largest prime?

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