10
$\begingroup$

As background, I don't know if this kind of calculations were in the literature and/or are interestings. I can to prove that being $\lambda\geq 1$ a fixed integer, $n$ is perfect if and only if $$2n=\left(\prod_{p\mid n}\frac{p^{e_p+1}-1}{p^{\lambda e_p+1}-1}\right)\sigma\left(n^\lambda\right),$$ where we suppose that $n$ has the factorization $n=\prod_{p\mid n}p^{e_p}$.

Inspired in the fact that it is easy to prove the following claim (and that simple cases as $\lambda=1,n=6$ or $\lambda=2,n=6$ don't work) I did a conjecture, that is my Question.

Claim. Let $n=\prod_{p\mid n}p^{e_p}$ an odd perfect number and as before $\lambda\geq 1$ a fixed integer, then $$\sigma(\xi)\prod_{p\mid n}\left(p^{e_p+1}-1\right)=\left(2^{\lambda+1}-1\right)2n\prod_{p\mid n}\left(p^{\lambda e_p+1}-1\right)$$ holds, where $\xi_{\lambda}=\xi=2^{\lambda}n^{\lambda}$.

Question. Let $n\geq 1$ an integer, and we take $\lambda\geq 1$ as a fixed integer. Prove or refute that if $n$ satisfies $$\sigma(\xi)\prod_{p\mid n}\left(p^{e_p+1}-1\right)=\left(2^{\lambda+1}-1\right)2n\prod_{p\mid n}\left(p^{\lambda e_p+1}-1\right),$$ where $\xi=2^{\lambda}n^{\lambda}$, then $n$ is an odd perfect number. Thanks.

My attempt (to get the statement as true). I know that the method is to prove by contradiction the statement on assumption that our $n$ has the form $2^{\alpha}m$ for integers $\alpha\geq 1$ and $m\geq 1$ with $(2,m)=1$. My deduction was then, if there are no typos, that $$\left(2^{\lambda(\alpha+1)+1}-1\right)\left(2^{\alpha+1}-1\right)\sigma(m)=\left(2^{\lambda+1}-1\right)2^{\alpha+1}m\left(2^{\lambda\alpha+1}-1\right)$$ in the way to do the comparison $\sigma(m)$ versus $\operatorname{something }\cdot m $, and to try deduce a contradiction, but I don't know how deduce it.

Then as motivation one could get such characterization for odd perfect numbers, that I am saying I don't know if is well known, if we can to finish the proof, for the veracity, of the statement.

$\endgroup$
3
+50
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Many thanks for your attention I believe that it is the idea, but I don't know how get it. That is, if there are no mistakes from my last equation then try bounding to deduce that it is a contradiction, as you said. $\endgroup$ – user243301 Jan 20 '17 at 13:42
  • $\begingroup$ @user243301, kindly check my updated answer to your original question. $\endgroup$ – Jose Arnaldo Bebita-Dris Jan 24 '17 at 14:57
  • 1
    $\begingroup$ Many thanks! I will need some hours. I didn't see before your expanded answer. Many thanks one more time for your attention. $\endgroup$ – user243301 Jan 24 '17 at 14:58
  • 1
    $\begingroup$ Sorry for thi late check as an accepted answer. Today I've read the full answer and understand the reasoning. $\endgroup$ – user243301 May 15 '17 at 10:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy