On odd perfect numbers $n$ and $\sigma\left(n^\lambda\right)$

Solution 1:

This is just a partial answer. One could get a contradiction by getting bounds on the quantity $I(m)=\sigma(m)/m$ based from your last equation, and restricting to the case $\alpha \geq 4$.

Expanded Answer

So you have $$I(m) = \dfrac{\sigma(m)}{m} = \dfrac{{2^{\alpha + 1}}(2^{\lambda\alpha + 1} - 1)(2^{\lambda + 1} - 1)}{(2^{\lambda(\alpha + 1) + 1} - 1)(2^{\alpha + 1} - 1)}.$$

This is bounded from below by $$L(\lambda, \alpha) := \dfrac{{2^{\alpha + 1}}(2^{\lambda\alpha + 1} - 1)(2^{\lambda + 1} - 1)}{(2^{\lambda(\alpha + 1) + 1})(2^{\alpha + 1})},$$ and is bounded from above by $$U(\lambda, \alpha) := \dfrac{{2^{\alpha + 1}}(2^{\lambda\alpha + 1})(2^{\lambda + 1})}{(2^{\lambda(\alpha + 1) + 1} - 1)(2^{\alpha + 1} - 1)}.$$ That is, we have $$L(\lambda, \alpha) < I(m) < U(\lambda, \alpha).$$

Simplifying and rewriting, we get $$L(\lambda, \alpha) = \dfrac{(2^{\lambda\alpha + 1} - 1)(2^{\lambda + 1} - 1)}{(2^{\lambda(\alpha + 1) + 1})} = 2\cdot\bigg(\dfrac{(2^{\lambda\alpha + 1} - 1)(2^{\lambda + 1} - 1)}{2^{\lambda(\alpha + 1) + 2}}\bigg)$$ $$= 2\cdot{\bigg(\dfrac{2^{\lambda\alpha + 1} - 1}{2^{\lambda\alpha + 1}}\cdot\dfrac{2^{\lambda + 1} - 1}{2^{\lambda + 1}}\bigg)},$$ and $$U(\lambda, \alpha) = \bigg(\dfrac{{2^{\alpha+1}}}{2^{\alpha+1}-1}\bigg)\cdot\bigg(\dfrac{2^{\lambda(\alpha + 1) + 2}}{2^{\lambda(\alpha + 1) + 1} - 1}\bigg) = 2\cdot\bigg(\dfrac{{2^{\alpha+1}}}{2^{\alpha+1}-1}\bigg)\cdot\bigg(\dfrac{2^{\lambda(\alpha + 1) + 1}}{2^{\lambda(\alpha + 1) + 1} - 1}\bigg).$$

By assumption, $\lambda \geq 1$ and $\alpha \geq 1$, so that we obtain the numerical bounds

$$\dfrac{9}{8} \leq L(\lambda, \alpha) < I(m) < U(\lambda, \alpha) \leq \dfrac{12}{7}.$$

Perhaps this argument could be tweaked so as to produce a contradiction? Please see additional notes below.

Added to Expanded Answer on January 24 2017

If $n=2^{\alpha}{m}$ (with $\alpha \geq 1$) and $n$ is perfect, then we get that $m = 2^{\alpha + 1} - 1$ is prime (so that $\alpha + 1$ is also prime), and we obtain $$I(m) = \dfrac{\sigma(m)}{m} = \dfrac{2^{\alpha + 1}}{2^{\alpha + 1} - 1}.$$

Now, if $\alpha \geq 4$, we get the upper bound $$I(m) \leq \dfrac{32}{31}$$ which contradicts the lower bound $$\dfrac{9}{8} < I(m).$$

It remains to consider the cases $n = 6$ and $n = 28$ (resp., $\alpha = 1$ and $\alpha = 2$) separately.