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(Thanks to Luis H Gallardo for pointing out the parity condition on $n$.)

(Edited on March 12, 2015)

I was actually trying to (initially) rule out the condition $\sigma(n) = q^k$ for an odd perfect number given in the Eulerian form $N = {q^k}{n^2}$. Just today, I found out that it might be possible to rule out the related condition $\sigma(n) = q$.

That being said, if we plug in $\sigma(n) = q^k$ into

$$2{n^2}\sigma(n)=\sigma(n^2)\sigma(\sigma(n))$$

with the (implicit) constraint $n = m^2$ (and since $n < \sigma(n) = q$ implies Sorli's conjecture that $k = 1$), then we get:

$$1 < I(q) \le \frac{6}{5} < \left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n) < I(n^2) < 2$$

where $I(x) = \displaystyle\frac{\sigma(x)}{x}$ is the abundancy index of $x$.

The lower bound $$\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n)$$ was communicated by Ochem to Dris in an e-mail dated April 17, 2013.

Note that $I(q^k)I(n^2) = 2$.

(That $\sqrt{I(n^2)} < I(n)$ follows from the inequality $$I(ab) \le I(a)I(b)$$ which is true $\forall a, b \in \mathbb{N}$. Equality holds if and only if $\gcd(a,b)$ = 1.)

Since $n = m^2$, $m$ divides $n$ and $m < n$, so that:

$$\sqrt{I(n)} = \sqrt{I(m^2)} < I(m) < I(m^2) = I(n)$$

But $\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n)$. Thus:

$$\sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} < I(m)$$

WolframAlpha gives the approximation $$\sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} \approx 1.20183425383797497745284556023594.$$

We therefore have $$I(q) \le \frac{6}{5} < \sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} < I(m).$$

Note that $\gcd(q,n)=\gcd(q,m)=1$, so that we have $q \neq m$. I now conjecture that $q < m = \sqrt{n}$ would follow from $I(q) < I(m)$ (and some further related inequalities), so that a contradiction against $n < \sigma(n) = q$ will arise.

(Thanks to Luis H Gallardo for pointing out the parity condition on $n$.)

(Edited on March 12, 2015)

I was actually trying to (initially) rule out the condition $\sigma(n) = q^k$ for an odd perfect number given in the Eulerian form $N = {q^k}{n^2}$. Just today, I found out that it might be possible to rule out the related condition $\sigma(n) = q$.

That being said, if we plug in $\sigma(n) = q$ into

$$2{n^2}\sigma(n)=\sigma(n^2)\sigma(\sigma(n))$$

with the (implicit) constraint $n = m^2$ (and since $n < \sigma(n) = q$ implies Sorli's conjecture that $k = 1$), then we get:

$$1 < I(q) \le \frac{6}{5} < \left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n) < I(n^2) < 2$$

where $I(x) = \displaystyle\frac{\sigma(x)}{x}$ is the abundancy index of $x$.

The lower bound $$\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n)$$ was communicated by Ochem to Dris in an e-mail dated April 17, 2013.

(That $\sqrt{I(n^2)} < I(n)$ follows from the inequality $$I(ab) \le I(a)I(b)$$ which is true $\forall a, b \in \mathbb{N}$. Equality holds if and only if $\gcd(a,b)$ = 1.)

Since $n = m^2$, $m$ divides $n$ and $m < n$, so that:

$$\sqrt{I(n)} = \sqrt{I(m^2)} < I(m) < I(m^2) = I(n)$$

But $\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n)$. Thus:

$$\sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} < I(m)$$

WolframAlpha gives the approximation $$\sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} \approx 1.20183425383797497745284556023594.$$

(Note that we have the equation $I(q^k)I(n^2) = I(q)I(n^2) = I(q)I(m^4) = 2$.)

We therefore have $$I(q) \le \frac{6}{5} < \sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} < I(m).$$

Note that $\gcd(q,n)=\gcd(q,m)=1$, so that we have $q \neq m$. I now conjecture that $q < m = \sqrt{n}$ would follow from $I(q) < I(m)$ (and some further related inequalities), so that a contradiction against $n < \sigma(n) = q$ will arise.

(Thanks to Luis H Gallardo for pointing out the parity condition on $n$.)

I was actually trying to rule out the condition $\sigma(n) = q^k$ for an odd perfect number given in the Eulerian form $N = {q^k}{n^2}$.

That being said, if we plug in $\sigma(n) = q^k$ into

$$2{n^2}\sigma(n)=\sigma(n^2)\sigma(\sigma(n))$$

with the (implicit) constraint $n = m^2$, then we get:

$$1 < I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < \sqrt{I(n^2)} < I(n) < I(n^2) < 2$$

where $I(x) = \displaystyle\frac{\sigma(x)}{x}$ is the abundancy index of $x$.

Note that $I(q^k)I(n^2) = 2$.

(That $\sqrt{I(n^2)} < I(n)$ follows from the inequality $$I(ab) \le I(a)I(b)$$ which is true $\forall a, b \in \mathbb{N}$. Equality holds if and only if $\gcd(a,b)$ = 1.)

Since $n = m^2$, $m$ divides $n$ and $m < n$, so that:

$$\sqrt{I(n)} = \sqrt{I(m^2)} < I(m) < I(m^2) = I(n)$$

But $\sqrt{\frac{8}{5}} < I(n)$. Thus:

$$\sqrt[4]{\frac{8}{5}} < I(m)$$

WolframAlpha gives the approximation $1.12468265$.

Unfortunately, as $$1 < I(q^k) < 1.25$$ I am now unable to make a comparison between the abundancy indices of $m = \sqrt{n}$ and $q^k$. I am confident, though, that by exploiting "symmetry", I will be able to get something more substantial in the coming days.

(Thanks to Luis H Gallardo for pointing out the parity condition on $n$.)

(Edited on March 12, 2015)

I was actually trying to (initially) rule out the condition $\sigma(n) = q^k$ for an odd perfect number given in the Eulerian form $N = {q^k}{n^2}$. Just today, I found out that it might be possible to rule out the related condition $\sigma(n) = q$.

That being said, if we plug in $\sigma(n) = q^k$ into

$$2{n^2}\sigma(n)=\sigma(n^2)\sigma(\sigma(n))$$

with the (implicit) constraint $n = m^2$ (and since $n < \sigma(n) = q$ implies Sorli's conjecture that $k = 1$), then we get:

$$1 < I(q) \le \frac{6}{5} < \left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n) < I(n^2) < 2$$

where $I(x) = \displaystyle\frac{\sigma(x)}{x}$ is the abundancy index of $x$.

The lower bound $$\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n)$$ was communicated by Ochem to Dris in an e-mail dated April 17, 2013.

Note that $I(q^k)I(n^2) = 2$.

(That $\sqrt{I(n^2)} < I(n)$ follows from the inequality $$I(ab) \le I(a)I(b)$$ which is true $\forall a, b \in \mathbb{N}$. Equality holds if and only if $\gcd(a,b)$ = 1.)

Since $n = m^2$, $m$ divides $n$ and $m < n$, so that:

$$\sqrt{I(n)} = \sqrt{I(m^2)} < I(m) < I(m^2) = I(n)$$

But $\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n)$. Thus:

$$\sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} < I(m)$$

WolframAlpha gives the approximation $$\sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} \approx 1.20183425383797497745284556023594.$$

We therefore have $$I(q) \le \frac{6}{5} < \sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} < I(m).$$

Note that $\gcd(q,n)=\gcd(q,m)=1$, so that we have $q \neq m$. I now conjecture that $q < m = \sqrt{n}$ would follow from $I(q) < I(m)$ (and some further related inequalities), so that a contradiction against $n < \sigma(n) = q$ will arise.

(Thanks to Luis H Gallardo for pointing out the parity condition on $n$.)

I was actually trying to rule out the condition $\sigma(n) = q^k$ for an odd perfect number given in the Eulerian form $N = {q^k}{n^2}$.

That being said, if we plug in $\sigma(n) = q^k$ into

$$2{n^2}\sigma(n)=\sigma(n^2)\sigma(\sigma(n))$$

with the (implicit) constraint $n = m^2$, then we get:

$$1 < I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < \sqrt{I(n^2)} < I(n) < I(n^2) < 2$$

where $I(x) = \displaystyle\frac{\sigma(x)}{x}$ is the abundancy index of $x$.

Note that $I(q^k)I(n^2) = 2$.

(That $\sqrt{I(n)} < I(n)$ follows from the inequality $$I(ab) \le I(a)I(b)$$ which is true $\forall a, b \in \mathbb{N}$. Equality holds if and only if $\gcd(a,b)$ = 1.)

Since $n = m^2$, $m$ divides $n$ and $m < n$, so that:

$$\sqrt{I(n)} = \sqrt{I(m^2)} < I(m) < I(m^2) = I(n)$$

But $\sqrt{\frac{8}{5}} < I(n)$. Thus:

$$\sqrt[4]{\frac{8}{5}} < I(m)$$

WolframAlpha gives the approximation $1.12468265$.

Unfortunately, as $$1 < I(q^k) < 1.25$$ I am now unable to make a comparison between the abundancy indices of $m = \sqrt{n}$ and $q^k$. I am confident, though, that by exploiting "symmetry", I will be able to get something more substantial in the coming days.

(Thanks to Luis H Gallardo for pointing out the parity condition on $n$.)

I was actually trying to rule out the condition $\sigma(n) = q^k$ for an odd perfect number given in the Eulerian form $N = {q^k}{n^2}$.

That being said, if we plug in $\sigma(n) = q^k$ into

$$2{n^2}\sigma(n)=\sigma(n^2)\sigma(\sigma(n))$$

with the (implicit) constraint $n = m^2$, then we get:

$$1 < I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < \sqrt{I(n^2)} < I(n) < I(n^2) < 2$$

where $I(x) = \displaystyle\frac{\sigma(x)}{x}$ is the abundancy index of $x$.

Note that $I(q^k)I(n^2) = 2$.

(That $\sqrt{I(n^2)} < I(n)$ follows from the inequality $$I(ab) \le I(a)I(b)$$ which is true $\forall a, b \in \mathbb{N}$. Equality holds if and only if $\gcd(a,b)$ = 1.)

Since $n = m^2$, $m$ divides $n$ and $m < n$, so that:

$$\sqrt{I(n)} = \sqrt{I(m^2)} < I(m) < I(m^2) = I(n)$$

But $\sqrt{\frac{8}{5}} < I(n)$. Thus:

$$\sqrt[4]{\frac{8}{5}} < I(m)$$

WolframAlpha gives the approximation $1.12468265$.

Unfortunately, as $$1 < I(q^k) < 1.25$$ I am now unable to make a comparison between the abundancy indices of $m = \sqrt{n}$ and $q^k$. I am confident, though, that by exploiting "symmetry", I will be able to get something more substantial in the coming days.