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
This contradicts:
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.
5 comments:
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
This contradicts:
Sigma(p)/m <= 2/3.
@Arnie Dris heaves a sigh of relief...@
Errors fixed... apologies to everyone... please have a look now... I have already double-checked everything... this should be correct now...
Re-formatted to improve the "visual" presentation...
Theorem is being retracted, for now...
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
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