As before, let N = (p^k)(m^2) be an Odd Perfect Number (OPN) with Euler prime p and gcd(p, m) = 1.
Earlier, we have seen that assuming the inequality m < p^k was true, the equality k = 1 would follow.
Just today (December 4, 2010), it seems as though (at least to myself) I am also able to prove that, by assuming the truth of the related inequality p^k < m, the same conclusion (i.e. the equality k = 1) follows.
These two implications together (of course) imply that it is indeed true that k = 1.
Thus, Sorli's conjecture on OPNs (circa 2003), which predicts that the exponent k of the Euler prime p should be unity (i.e. k = 1), can now be rightfully called a theorem.
I will be presenting the details of the proof by December 6 (Monday) [early afternoon] to a group of mathematicians (with particular specializations in algebraic number theory, elliptic curves and coding theory) at the University of the Philippines - Diliman.
Some random thoughts on pseudo-random things
1 comment:
Proof has been clarified (just today).
While k = 1 does NOT follow from p^k < m (at least, using my methods), the two mutually exclusive conditions:
(1) k = 1, m < p^k
(2) k > 1, p^k < m
do give a complete characterization for OPNs. Indeed, over MathOverflow I was able to show that the first condition would imply that there are no OPNs (i.e. Sorli's conjecture implies the OPN conjecture).
Thus, the only remaining case to tackle at this point (in order to prove the OPN conjecture in full generality) will be the case when k > 1 and p^k < m.
See these posts for related information:
http://arnienumbers.blogspot.com/2010/12/proof-for-sorlis-conjecture-on-odd.html
http://mathoverflow.net/questions/48203/on-sorlis-conjecture-re-opns-circa-2003
Post a Comment