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Let $\sigma(x) = \sigma_1(x)$ denote the sum of all the positive divisors of $x$.

If $n \in \mathbb{N}$ is odd and $\gcd(n, \sigma(n)) = 1$, then do there exist any solutions to the following equation?

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

In other words, does there exist such an odd $N = {n^2}\sigma(n)$ which is also a perfect number?

Here, I also list some (open?) problems mentioned by other researchers:

(1) Suryanarayana: Is it true that every odd perfect number is of the form $m\sigma(m)$ for some odd integer $m$; if so, is $\gcd(m, \sigma(m)) = 1$ necessarily?

(2) M. V. Subbarao: Does every odd perfect number $n$ (if such exist) have the representation $$n = \frac{1}{2}m\sigma(m) \hspace{0.5in} (*)$$

Another question: Whenever $n$ given by $(*)$ is perfect, does it follow that $n$ is odd and $\gcd(m, \sigma(m)) = 1$?

I would appreciate it if anybody could point me to reference(s) to recent results on either Problem (1) or Problem (2).

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recent reference of what? –  Franz Lemmermeyer Mar 2 '11 at 19:42
    
@Franz, by "recent reference" I meant to refer to any new results in the direction of resolving (1) or (2). Editing my question to reflect this. Thanks! –  Jose Arnaldo Bebita Dris Mar 2 '11 at 19:51
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You would do better by searching arXiv.org and other sources for information on multiperfect and odd perfect numbers, and then asking for a reference request for ONE SPECIFIC PROBLEM if you were worried that you had missed something. If he has the time, I bet Pace Nielsen could more than satisfactorily fill a modest reference request. Gerhard "Baby Don't Let Me Down!" Paseman, 2011.03.02 –  Gerhard Paseman Mar 2 '11 at 21:11
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In order to have $N=n^2 \sigma(n)$ odd you need $$ n = k^2 $$ (since you want $n$ odd also) for some positive integer $k$, so that you ask for a possible odd perfect number of the form $$ N =k^4 \sigma(k^2) $$ To find any possible odd perfect number seems difficult; one of this more particular form probably is still more difficult to find... –  Luis H Gallardo Mar 2 '11 at 21:52
    
Thank you very much Luis! I didn't realize earlier that parity conditions would settle my original problem. =) Coming up with an answer now. –  Jose Arnaldo Bebita Dris Mar 3 '11 at 4:41

1 Answer 1

up vote 1 down vote accepted

(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.

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As you can see, this method can be likened to a form of descent, coupled with exploiting symmetry via the divisibility constraint $\gcd(q, n) = 1$. More on this soon. –  Jose Arnaldo Bebita Dris Mar 3 '11 at 5:31

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