Let N = q^k * n^2 be an odd perfect number, with q prime,
q == k == 1 (mod 4) and gcd(q, n) = 1.
We wish to show the following equivalence:
Theorem: k > 1 if and only if q^k < sigma(q^k) < n < sigma(n)
We do this by first proving the following lemmas:
Lemma 1: If n < q <= q^k, then k = 1.
Suppose n < q.
In my MS thesis, I proved that q^k < n^2.
Since n < q, then n^2 < q^2.
Therefore, n < q <= q^k < n^2 < q^2, which implies
that 1 <= k < 2.
Since k == 1 (mod 4), it follows that k = 1.
Remark 1: The contrapositive of Lemma 1 is:
If k > 1, then q <= q^k < n.
Lemma 2: If k = 1, then n < q <= q^k.
Suppose k = 1. Assume to the contrary that q^k < n.
Then q^1 = q < n. Thus, k = 1 implies q < n.
The contrapositive of the last statement is
n < q implies k > 1. This contradicts Lemma 1.
Edit (January 28, 2012): n < q implies k > 1 does not
contradict Lemma 1. This is because we do not know
if n < q is indeed true.
Remark 2: By Lemma 1 and Lemma 2, we now know
that k = 1 if and only if n < q <= q^k.
Lemma 3: sigma(q^k) >= sigma(q) > n if and only if q > n.
Suppose that sigma(q) > n. Assume to the contrary that n > q.
Then we have sigma(q) = q + 1 > n > q, contradicting the fact
that n is a natural number.
Suppose that q > n. Since sigma(q) > q, we have
sigma(q) > q > n and we are done.
Since sigma(q^k) >= sigma(q) for all k, then the lemma is proved.
We now have the following claim:
Claim: k > 1 if and only if q^k < sigma(q^k) < n < sigma(n)
Proof of Claim:
It suffices to show that k > 1 if and only if sigma(q^k) < n.
By Lemma 1, Lemma 2 and Lemma 3, k = 1 if and only if
n < q <= q^k if and only if sigma(q^k) >= sigma(q) > n. By
k > 1 if and only if q <= q^k < sigma(q^k) < n < sigma(n)
The claim follows. And therefore, the theorem mentioned at the
beginning of this post is proved.
Via the contrapositive:
k = 1 if and only if n < sigma(n) <= q^k < sigma(q^k)
I believe that this last result will now enable me to rule out
Case B.1 as mentioned in pages 16 through 17 of my paper,