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.

Proof 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.**

**Proof 2:**

**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.

Proof 3:

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

the contrapositive:

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:

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,

hyperlinked here.