# On Sorli's Conjecture Re: OPNs (Circa 2003)

In the PhD dissertation titled "Algorithms in the Study of Multiperfect and Odd Perfect Numbers" (hyperlinked here) and completed in 2003, Ronald Sorli conjectured that the exponent $k$ on the Euler prime $p$ for an odd perfect number $N = (p^k)(m^2)$ is one (i.e. we can drop $k$).

Assuming Sorli's conjecture is true, does anyone know if there exist (any) "effective" results (pardon my use of the term, I just could not think of a better word) in the literature, particularly with respect to relations between the Euler prime $p$, the exponent $k$ and the number $\sqrt{\frac{N}{p^k}}$? I have, so far, only been able to get hold of Paolo Starni's article titled "Odd Perfect Numbers: A Divisor Related to the Euler′s Factor".

In particular, note that Sorli's conjecture implies the following relations:

$$I(p^k) = I(p) = \frac{p+1}{p}$$

$$I(m^2) = \frac{2}{I(p)} = \frac{2p}{p + 1}$$

which, in turn, gives the (trivial) algebraic identity:

$$\frac{1}{p} = \frac{1}{p+1} + \frac{1}{p}\left(\frac{1}{p+1}\right)$$

where $p$ is the Euler prime (i.e. $p^k$ is the Euler's Factor) and $$I(x) = \frac{\sigma(x)}{x}$$ is the abundancy index of $x$.

[Edit (September 18 2013) - Per Professor Beasley's paper titled "Euler and the Ongoing Search for Odd Perfect Numbers" from this hyperlink:

Before proceeding with Euler’s proof, we pause to note that his result was not quite what Descartes and Frenicle had conjectured, as they believed that $k = 1$, but it came very close. In fact, current research continues in an effort to prove $k = 1$. For example, Dris has made progress in this direction, although his paper refers to Descartes’ and Frenicle’s claim (that $k = 1$) as Sorli’s conjecture; Dickson has documented Descartes’s conjecture as occurring in a letter to Marin Mersenne in 1638, with Frenicle’s subsequent observation occurring in 1657.

End Edit.]

• Consider the sum $I(p^k)+I(m^2)$. Then we have the (sharp) bounds $L(p)=\frac{3p^2−4p+2}{p(p−1)}$ and $U(p)=\frac{3p^2+2p+1}{p(p+1)}$ with $L(p) \lt I(p^k)+I(m^2) \le U(p)$. If Sorli's conjecture is proved, then $I(p^k)+I(m^2)=U(p)$. Commented Dec 5, 2010 at 7:02
• Additionally, the derivative $U'(p) \gt 0$ and, while $U(p)$ has no maximum value on the interval $[5, \infty)$, it does have a least upper bound of $\lim_{p\to\infty}{U(p)} = 3$. As remarked by Joshua Zelinksy a few years back: "Any improvement on the upper bound of $3$ would have similar implications for all arbitrarily large primes and thus would be a very major result." (e.g. $U(p) < 2.99$ implies $p \le 97$.) Thus, (assuming Sorli's conjecture), "heuristically" we can have the following approach for the OPN problem: Fix an upper bound for $u = U(p)$, and then use factor chains... Commented Dec 5, 2010 at 7:30
• ... to loop until you get the contradiction $U^{-1}(p) < p$. Commented Dec 5, 2010 at 7:33
• Update: If Sorli's conjecture is indeed true, then there are no odd perfect numbers. The proof proceeds via reductio ad absurdum, and is replete with all sorts of contradictions at every (succeeding) step. This reminds me of James Joseph Sylvester's quote from 1888: "… a prolonged meditation on the subject has satisfied me that the existence of any one such [OPN] — its escape, so to say, from the complex web of conditions which hem it in on all sides — would be little short of a miracle." Commented Dec 7, 2010 at 14:18
• Arnie, that is a bold claim. You really should back it up. To claim that if $k=1$ there are no OPN's would be a very big deal. Commented Dec 7, 2010 at 17:49

As far as I know, there are no such effective bounds. In fact, even if $p=5$ and $k=1$, there are no known effective bounds on $N$. (There are bounds on $N$ in terms of the number of distinct factors.)
• @Pace, thanks a lot! What about if we consider the sum $$I(p^k) + I(m^2)$$ with this additional information from Sorli's conjecture? Can you comment on that one? Commented Dec 4, 2010 at 4:18
• I usually think of I as a multiplicative function. But I imagine that you can use the identities you derived above to get a lower bound on this thing. But again, if $p=5$ and $k=1$ this sum is just $6/5 + 5/3$, and there isn't much to say. Commented Dec 6, 2010 at 23:36