## 9.12.11

### Conditions Equivalent to Sorli's Conjecture Re: Odd Perfect Numbers

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.
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,