First, we prove the following result:

**Lemma 1.** Let $N = {q^k}{n^2}$ be an odd perfect number (OPN) given in Eulerian form. If $k = 1$ implies $\sigma(n) < q^k$, then $k = 1$ if and only if $n < q$.

**Proof.**

Suppose $N = {q^k}{n^2}$ is an odd perfect number (OPN) given in Eulerian form.

If $k = 1$ implies $\sigma(n) < q^k$, then we have:

$k = 1 \Longrightarrow \sigma(n) < q^k \Longrightarrow n < \sigma(n) < q^k = q.$

Thus, we have: $k = 1 \Longrightarrow n < q$.

But we also have: $n < q \Longrightarrow k = 1$ [

Dris, 2012].

Therefore, $k = 1$ if and only if $n < q$.

**QED.**

**Remark 1.** Notice that, in fact if we have $n < q^k$, then $k = 1 \Longleftrightarrow n < q$ would follow. Furthermore, note that if $n < q$, then we have $n < q^k$ (since it is true that $k \geq 1$, in general). Consequently, to prove the biconditional $k = 1 \Longleftrightarrow n < q$, it suffices to prove the inequality $n < q$.

Next, we prove the following proposition:

**Lemma 2.** Let $N = {q^k}{n^2}$ be an odd perfect number (OPN) given in Eulerian form. If $k = 1$ implies $\sigma(q^k) < n$, then $q < n$.

**Proof.**

Suppose $N = {q^k}{n^2}$ is an odd perfect number (OPN) given in Eulerian form.

If $k = 1$ implies $\sigma(q^k) < n$, then we have:

$k = 1 \Longrightarrow \sigma(q^k) < n \Longrightarrow q < \sigma(q) = \sigma(q^k) < n$.

Therefore, we have: $k = 1 \Longrightarrow q < n$.

But we also have: $n < q \Longrightarrow k = 1$ [

Dris, 2012]. By the contrapositive:

$k \neq 1 \Longrightarrow q < n$.

Consequently, we conclude that $q < n$.

**QED.**

**Remark 2.** If $\sigma(q^k) < n$, then $q^k < n$. (This last inequality was conjectured in the M.Sc. thesis [

Dris, 2008] and in the published article [

Dris, 2012]). Note that, if $q^k < n$, then it is trivial to prove that $k \geq 1 \Longleftrightarrow q < n$.

*This*** blog post is currently a WORK IN PROGRESS. Please do check back from time to time for updates!**