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(I have asked a similar question in MSE four days ago, but did not receive any answers. I have therefore cross-posted it to this site, hoping to get some responses.)

An odd perfect number $N$ is said to be given in Eulerian form if $N = {q^k}{n^2}$ where $q$ is prime, $q \equiv k \equiv 1 \pmod 4$, and $\gcd(q, n) = 1$.

Here is my question:

If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, is $n$ a square?

I would certainly appreciate it if someone could point me to existing papers in the literature where this particular question is addressed.

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  • $\begingroup$ if an odd superfect number exist they are square , just to check this paper :G. G. dandapat, J. L. Sunsucker, C. Pomerance, Some new results on odd perfect numbers, Paci¯c J. Math., 57 (1975), 359{364., and if you take n=p which is prime you will accross one of the open conjecture about :odd hyperperfect number and try to check this paper:J. S. McCranie, A study of hyperperfect numbers, J. Integer Seq., 3 (2000), Article 00.1.3. $\endgroup$
    – zeraoulia rafik
    Commented Dec 31, 2015 at 14:19
  • $\begingroup$ @zeraouliarafik, I was aware of the first paper you mention even before posting this question. However, that covers the case of odd $M$ for which $\sigma(\sigma(M)) = 2M$, while I am asking for the case of odd $N$ for which $\sigma(N) = 2N$. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Dec 31, 2015 at 14:22
  • $\begingroup$ The existence of superperfect number implies the existence of odd perfect number . $\endgroup$
    – zeraoulia rafik
    Commented Dec 31, 2015 at 14:25
  • $\begingroup$ @zeraouliarafik, well yes! But does the fact that odd superperfect numbers are square also imply that the square root of the non-Euler part $n^2$ of an odd perfect number given in the Eulerian form $N = q^k n^2$, must also be a square? That is essentially what I am asking for in this question. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Dec 31, 2015 at 14:29
  • $\begingroup$ Per McCranie's JIS paper: "A number $n$ is $k$-hyperperfect for some integer $k$ if $n = 1 + k(\sigma(n) - n - 1)$. The $1$-hyperperfect numbers are the familiar perfect numbers." $\endgroup$
    – Jose Arnaldo Bebita
    Commented Dec 31, 2015 at 14:32

1 Answer 1

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There is no reason to believe that the non-Eulerian part is a 4th power. Moreover, Descartes spoof OPN shows us that such a guess is probably unmotivated and that there is no purely combinatorial way to prove that $n$ is a square. A proof of such a result would fundamentally require restriction to actual (rather than spoof) prime factors; and as far as I'm aware most results on OPN's (besides the finite computational searches) don't make such a restriction.

However, it may be possible to prove that $n$ cannot be a perfect square. It has been long known that it cannot happen that $p^2||n$ for each prime $p|n$. For a vast generalization of this result see my paper with S. Adam Fletcher and Pascal Ochem: Sieve methods of odd perfect numbers.

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