## 25.8.12

### OPN Research - August 2012 - Part 4

Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form (i.e. $q \equiv k \equiv 1 \pmod 4$) and $\gcd(q, n) = 1$).

Recall that Sorli's conjecture (made in the year 2003) predicts that $k = 1$.

From a previous blog post, we have the following result:

Lemma 1:  If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then $k = 1$ and $q < n$ imply that $q^k < n$.

Proof of Lemma 1:  Trivial.

Remark 1:  The contrapositive of the implication in Lemma 1 is as follows:

$n < q^k$ implies $k > 1$ or $n < q$.

Now, suppose that $n < q^k$.

If $n < q$, then $k = 1$.  (This is essentially due to Lemma 3.1, page 8 in this arXiv preprint.)

If $k = 1$, then $n < q$.  (Trivial.)

Thus, if $n < q^k$, then $k = 1$ if and only if $n < q$.

Additionally, if $n < q^k$, then $k > 1$ if and only if $q < n$.

We state the immediately preceding results as our second lemma:

Lemma 2:  Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form.  If $n < q^k$, then $k = 1$ if and only if $n < q$.

Going back to Lemma 1:

$k = 1$ and $q < n$  imply that $q^k < n$.

We want to establish results in the other direction.

To this end, suppose $q^k < n$.

If $k \geq 1$, then $q < n$. (This can be seen by writing the inequalities as $q \leq q^k < n$.)

If $q < n$, then $k \geq 1$. (This implication is true because $q < n$ is true [it follows from our assumption $q^k < n$] and $k \geq 1$ is also true [since $k \equiv 1 \pmod 4$].)

Thus, $k \geq 1$ if and only if $q < n$.

We summarize the results in the preceding paragraphs in our third lemma:

Lemma 3: Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form.  If $q^k < n$, then $k \geq 1$ if and only if $q < n$.

Remark 2:  Note the contrast between

Lemma 2:  If $n < q^k$, then $k = 1$ if and only if $n < q$.
and
Lemma 3:  If $q^k < n$, then $k \geq 1$ if and only if $q < n$.

Note that, if we could rule out the condition "$k = 1$" in the biconditional in Lemma 3's conclusion, then essentially we would be able to come up with the following "result":

Conjecture C:  Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form.  Then $k = 1$ if and only if $n < q$.

Remark 3:  If Conjecture C is proved, then a subsequent proof for the inequality $q < n$ or the stronger inequality $q^k < n$ (the latter of which was conjectured by the author in his M. Sc. thesis, completed in the year 2008) would then disprove Sorli's conjecture regarding odd perfect numbers. In Pace Nielsen's opinion (expressed over MathOverflow, in a comment to the author's question hyperlinked here), this "would be a very big deal".