## 5.11.10

### NOT a Proof - BUT Some Partial Results for the Odd Perfect Number Conjecture

D.   Bounds for the Abundancy Indices of (p^k) and (m^2) when N = (p^k)(m^2) is an OPN
with Euler prime p

(#) 1 < I(p^k) < 5/4 < 8/5 < I(m^2) < 2 < 57/20 < I(p^k) + I(m^2) < 3

($) 0 < (p^k)/(m^2)I(p^k) <= 2/3 < 3 <= (m^2)/(p^k)I(m^2) < (p^k)/(m^2)I(p^k) + (m^2)/(p^k)I(m^2) (%) 0 < 1 < I(p^k) < 5/4 < 2*SQRT(5)/10 < I(p^k)I(m) < 2 (&) 11/3 <= (p^k)/(m^2)I(p^k) + (m^2)/(p^k)I(m^2) The order relations ($), (%) and (&) were all derived from the following two observations:

(1) Divisibility/Congruence Tricks

gcd(p^k, Sigma(p^k) = 1 implies p^k divides Sigma(m^2). This means that Sigma(m^2)/p^k is an integer.

Additionally, since squares are never perfect, and also because m^2 is a divisor of N which is perfect (and hence, m^2 is deficient), therefore:

Sigma(m^2) is always odd. Now, p^k is an odd prime power (since N is odd). Thus, Sigma(m^2)/p^k is odd.

Let h := Sigma(m^2)/p^k. Suppose h = 1.

Because of the relation Sigma(p^k)Sigma(m^2) = 2(p^k)(m^2), we have:

2/h = Sigma(p^k)/m^2

Thus, we have the simultaneous Sigma relations:

(*) Sigma(p^k) = 2m^2

(**) Sigma(m^2) = p^k

However, according to the paper titled “Some New Results on Odd Perfect Numbers” by G. G. Dandapat,
J. L. Hunsucker and Carl Pomerance: No OPN satisfies both (*) and (**).

Thus, h = Sigma(m^2)/p^k = [(m^2)/(p^k)]I(m^2) >= 3.

(2) Inequality/Numerical Tricks

gcd(m, n) = 1 if and only if I(m)I(n) = I(mn)

What happens when equality does not hold?

We have: I(mn) < I(m)I(n)

In particular, when m = n, then we have:

I(m^2) < [I(m)]^2

But from the appropriate order relation in (#) we get:

8/5 < [I(m)]^2

Taking square roots:

2*SQRT(10)/5 < I(m) (Note that: SQRT(10) is approximately 3.162277660168379, which is bigger than, yet surprisingly proximate, to PI.)

Computing 2*SQRT(10)/5, this is approximately 1.2649110640673515.

But I(p^k) < 5/4 = 1.25.

Thus, we have I(p^k) < I(m). (Please do note that it was ALMOST A MISS.)

Consider the equation Sigma(p^k) = m.

Sigma(p^k) = 1 + p + p^2 + ... + p^k == 1 + 1 + 1 + ... + 1 (mod 4)
since the Euler prime p == 1 (mod 4)
== k + 1 (mod 4) == 2 mod (4) since k == 1 (mod 4)

Thus m == 2 mod 4, a contradiction to the assumption that N is an OPN.

Nov 7, 2010 : (The equation I(p^k) = m is not equivalent to
the equation Sigma(p^k) = m.  However, a simpler
argument will be to consider that if p = m, then

N = (p^k)(m^2) = (p^(k + 2)),

a prime-power and which is known to be deficient,
whereas we assumed N is an OPN.)

Consequently, we have two cases for further study:

Case 1: Sigma(p^k) < m

Case 2: m < Sigma(p^k)

Let's consider Case 2 first.

Since m < Sigma(p^k) implies m < (5/4)p^k, let's consider a "mathematical model for OPN(s)" that is simpler and better:

Since prime powers are always deficient, m is not equal to p. Consider Case 2 in light of this observation.  ( e.g. Suppose m < p ( <= p^k < Sigma(p^k) ). )

Since (p^k)/(m^2)I(p^k) <= 2/3 and I(p^k) > 1, then this means that:

(p^k)/(m^2) < 2/3 < 1.

In particular, p <= p^k < (2/3)(m^2) < (2/3)p^2 < p^2.

That is, p^(1) <= p^(k) < p^2.

Consequently, 1 <= k < 2.

In particular, since we know a priori that k must be congruent to 1 modulo 4, it follows that k = 1.

It follows that, if m < p^k and N = (p^k)(m^2) is an OPN, then k = 1 (e.g. p || N).  (This was conjectured by Ronald Sorli in 2003.)  We will show that this will then lead to a contradiction.

We are assuming that m < p ( and of course, p <= p^(k) ).

But I(p) = Sigma(p)/p, which means that:

m < p < Sigma(p) < Sigma(m) < 2m

since p and m are deficient.

But we also know that:

Sigma(p)/m <= 2/3.  --> Retracted.

What we do know is that:

[Sigma(p)]/(m^2) <= [Sigma(p^k)]/(m^2) = [(p^k)/(m^2)]I(p^k) <= 2/3.

Of course, since [Sigma(p)]/(m^2) < [Sigma(p)]/m, the required conclusion does NOT follow.

Thus, we have:

m < p < Sigma(p) < Sigma(m) < 2m

which implies that:

m < Sigma(p) < 2m

from which it follows that:

1 < Sigma(p)/m < 2

Sigma(p)/m <= 2/3.

This leaves us to consider Case 1: Sigma(p^k) < m (in light of the
simplified “mathematical model” p < m.)

Assume Case 1 is true. Then it would imply that (p^k)/(m) < 1. Equivalently, (m)/(p^k) > 1.

Therefore, we now have the following theorem:

Theorem [Arnie Dris, 2010]: If N = {p^(k)}{m^(2)} is an OPN, then p^(k) < m. --> Retracted.

PostScript:
When I completed my masters thesis in 2008 (it was titled "Solving the Odd Perfect Number Problem - Some Old and New Approaches"), I conjectured that:

p^k < m

on the basis of the *very limited* information that p^k < m^2 "might have" followed from I(p^k) < I(m^2). Indeed, I was under the naive assumption that the "divisibility constraint" gcd(p, m) = 1 together with I(p^k) < I(m^2) induced some sort of "order property" between p^k and m^2.

Arnie Dris said...

Oops, my bad. There was a serious error there.

It is indeed true that:

m < p

But I(p) = Sigma(p)/p, which means that:

m < p < Sigma(p) < Sigma(m) < 2m

since p and m are deficient.

But we also know that:

Sigma(p)/m <= 2/3.

But:

m < p < Sigma(p) < Sigma(m) < 2m

implies that:

m < Sigma(p) < 2m

from which it follows that:

1 < Sigma(p)/m < 2

Sigma(p)/m <= 2/3.

@Arnie Dris heaves a sigh of relief...@

Arnie Dris said...

Errors fixed... apologies to everyone... please have a look now... I have already double-checked everything... this should be correct now...

Arnie Dris said...

Re-formatted to improve the "visual" presentation...

Arnie Dris said...

Theorem is being retracted, for now...

Arnie Dris said...

Error in the indicated line:

But we also know that:

Sigma(p)/m <= 2/3. <-----

Compare with what is known:

Sigma(p)/m^2 <= 2/3