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8.12.10

Are the Euler Primes in One-to-One Correspondence with the Odd Perfect Numbers?

Here we consider the problem of establishing a one-to-one correspondence between "the" Euler primes and the Odd Perfect Numbers (OPNs), in the same way that it is (currently) known that the Mersenne primes are in one-to-one correspondence with the Even Perfect Numbers (EPNs).

Let us recapitulate what we have been able to prove so far:

(A)  Sorli's conjecture implies the OPN conjecture, provided we could rule out the specific condition

p = q = 5

as detailed out in this post.  (Note that p = 5 is the smallest possible Euler prime since p == 1 (mod 4).)

(B)  The abundancy index value of 2 (i.e. the classical concept of number perfection) provides a "natural bifurcation" between EPNs and OPNs, in the following sense:

(1)  Mersenne primes are congruent to 3 modulo 4, while "the" Euler primes are congruent to 1 modulo 4.
(2)  If we can prove (A) (i.e. by ruling out p = q = 5), then we will be able to show the following (full) characterization theorem for perfect numbers:

"Enhanced Euclid-Euler Model for Perfect Numbers":
If M = (p^a)(n^2) is a perfect number with gcd(p, n) = 1, then
(a)  M is even if and only if a = 1 and n < p^a.
(b)  M is odd if and only a > 1 and p^a < n.

Finally, I give you some inequality plots (again from WolframAlpha) which I hope will clarify why it cannot happen that p = q = 5 (and therefore why p is different from q, in general):

Inequality Plot for (p - 1)(q - 1) < (p + 1)(q - 2), 2q >= p + 5, q = p




Inequality Plot for (p - 1)(q - 1) < (p + 1)(q - 2), 2q >= p + 5, q > p



Inequality Plot for (p - 1)(q - 1) < (p + 1)(q - 2), 2q >= p + 5, p < q




Inequality Plot for (p - 1)(q - 1) < (p + 1)(q - 2), 2q >= p + 5



Inequality Plot for (p - 1)(q - 1) < (p + 1)(q - 2), q = p



Inequality Plot for (p - 1)(q - 1) < (p + 1)(q - 2), q > p



Inequality Plot for (p - 1)(q - 1) < (p + 1)(q - 2), q < p




Inequality Plot for (p - 1)(q - 1) < (p + 1)(q - 2)



Again, I am "heuristically" claiming that:

CONJECTURE I:  There is a one-to-one correspondence between the Euler primes and the OPNs.

In effect, Conjecture I is equivalent to the following related (albeit more general) conjecture:

CONJECTURE II:  All squares are solitary numbers.

Please see these links for an attempt to prove Conjecture II:

[1]  http://groups.google.com/group/sci.math/msg/c9901d8a2e03e8f4

[2]  http://groups.google.com/group/sci.math/browse_thread/thread/a68aa1190a8dbbd7/c9901d8a2e03e8f4#c9901d8a2e03e8f4


These two links are archived in:

[3]  http://upforthecount.com/math/abundance.html

You might also be interested in:

[4]  http://upforthecount.com/math/perfect.html


In particular, note from [1] and [2] that an initial check done by Robert Israel (from Vancouver, Canada) using Maple did not produce any counterexamples to conjecture II in the range

1 <= A < B <= 300000

where

I(A^2) = Sigma(A^2)/A^2 == Sigma(B^2)/B^2 = I(B^2).


I finished typing this past 10:30 PM (Manila time) [last edited at 1:05 AM].  That's all for now.  Will try to resume working on clarifying the salient points by tomorrow.

2 comments:

Arnie Dris said...

Conjecture I has been proved. No need to resort to a proof of the more general Conjecture II. I believe this last detail clarifies everything that Pace Nielsen needed to know.

Arnie Dris said...

In addition, I believe I can now prove the OPN Conjecture in full generality, even without having to bother with Case 2. I am writing a blog post in that direction now.