Does there exist a nontrivial prime power $q^k$ such that $\sigma(n^2)/n = q^k$ for some $n$?

Question

Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of the positive integer $x$.

My question in the present post is closely related to this one in MO:

QUESTION

Does there exist a nontrivial prime power $q^k$ such that $\sigma(n^2)/n = q^k$ for some $n$?

I tried checking for examples of numbers $2 \leq n \leq {10}^6$ satisfying the divisibility constraint $$n \mid \sigma(n^2)$$ using a Pari-GP script, via Sage Cell Server:

for(n=2, 1000000, if((Mod(sigma(n^2),n) == 0),print(n,factor(n))))

Here is the output:

39[3, 1; 13, 1]
793[13, 1; 61, 1]
2379[3, 1; 13, 1; 61, 1]
7137[3, 2; 13, 1; 61, 1]
13167[3, 2; 7, 1; 11, 1; 19, 1]
76921[13, 1; 61, 1; 97, 1]
78507[3, 2; 11, 1; 13, 1; 61, 1]
230763[3, 1; 13, 1; 61, 1; 97, 1]
238887[3, 2; 11, 1; 19, 1; 127, 1]
549549[3, 2; 7, 1; 11, 1; 13, 1; 61, 1]
692289[3, 2; 13, 1; 61, 1; 97, 1]
863577[3, 2; 11, 2; 13, 1; 61, 1]

The Pari-GP interpreter of Sage Cell Server crashes as soon as a search limit of ${10}^7$ is specified.

OEIS sequence A232354 lists $187$ additional examples.

I skimmed through the list of the first $199$ examples (not including $1$) in OEIS and noted that all of them are odd. Additionally, here are the corresponding integer values for $\sigma(n^2)/n$ for $2 \leq n \leq {10}^6$: $$\frac{\sigma({39}^2)}{39} = 61,$$ $$\frac{\sigma({793}^2)}{793} = 873 = {3^2} \times {97},$$ $$\frac{\sigma({2379}^2)}{2379} = 3783 = 3 \times {13} \times {97},$$ $$\frac{\sigma({7137}^2)}{7137} = 11737 = {11}^2 \times {97},$$ $$\frac{\sigma({13167}^2)}{13167} = 26543 = {11} \times {19} \times {127},$$ $$\frac{\sigma({76921}^2)}{76921} = 85563 = {3^3} \times {3169},$$ $$\frac{\sigma({78507}^2)}{78507} = 141911 = 7 \times {11} \times {19} \times {97},$$ $$\frac{\sigma({230763}^2)}{230763} = 370773 = {3^2} \times {13} \times {3169},$$ $$\frac{\sigma({238887}^2)}{238887} = 417263 = 7 \times {11} \times {5419},$$ $$\frac{\sigma({549549}^2)}{549549} = 1155561 = 3 \times {11} \times {{19}^2} \times {97},$$ $$\frac{\sigma({692289}^2)}{692289} = 1150347 = 3 \times {{11}^2} \times {3169},$$ $$\frac{\sigma({863577}^2)}{863577} = 1562185 = 5 \times {97} \times {3221}.$$

(I did the computations one at a time using WolframAlpha.)

Note that $61$ is prime. However, it is not a nontrivial prime power.

I was hoping somebody with more computing power (and better programming skills) could automate the computation and thereby yield my desired result, if there is one at all. (I would be content if this computation were done over this list in OEIS.)

Thank you!

This inquiry is also related to this answer to a closely related question, which currently has a pending bounty. — Jose Arnaldo Bebita Dris, Feb 06 '22 at 13:09
That is a huge search limit ! If no solution is within this search limit, I do not expect a solution at all. Let us think about useful restrictions for $n$ ... — Peter, Feb 06 '22 at 13:51
My computations above are only for $2 \leq n \leq {10}^6$, @Peter. Apparently, the closest solution that we could get to $\sigma(n^2)/n = q^k$ is only when $q = 61$ and $k = 1$. But then, we require $k > 1$ in order for $q^k$ to be a nontrivial prime power. So I conjecture that there might be no solutions to my computational search problem. Alas, I have no proof! T_T — Jose Arnaldo Bebita Dris, Feb 06 '22 at 14:26
I checked further and manually computed and then factored the relevant quotients $\sigma(n^2)/n$ (using WolframAlpha) for ${10}^6 < n \leq {10}^8$, and did not find any solutions, @Peter. (No further instance with $q$ prime and $k = 1$.) — Jose Arnaldo Bebita Dris, Feb 06 '22 at 15:14
My routine needed quite long for the humble range upto $n=10^8$ , no integer which is a perfect power appeared here, I did not doublecheck whether another prime occurs. We need criterions ($n$ must be odd but this does not help much) which I think the OEIS-author used as well to accelerate the search. — Peter, Feb 06 '22 at 15:38
There is already a list of the first 199 $n$ such that $2 \leq n \leq {6.8} \times {10}^{12}$ here. We only need to check for nontrivial prime-power values of $\sigma(n^2)/n$ for such $n$ in that list, @Peter. — Jose Arnaldo Bebita Dris, Feb 07 '22 at 00:09
I mean, I would be content with a check done in that range, over that list, @Peter. — Jose Arnaldo Bebita Dris, Feb 07 '22 at 00:10
Then it looks more like programming question (for stack overflow?)... Anyway I have loaded the list into Maple and checked that only $n=39$ there leads to single $p^e$ component. — Sil, Feb 09 '22 at 12:42
@Sil: For the record, $n=39$ does lead to $\sigma(n^2)/n = p^e$, where $p=61$ and $e=1$. Thank you very much for your time and attention! =) — Jose Arnaldo Bebita Dris, Feb 09 '22 at 23:53
@Sil: Please flesh out your comment into an actual answer, so that I could upvote. Thanks! — Jose Arnaldo Bebita Dris, Feb 11 '22 at 09:15

Sil · Answer 1 · 2022-02-11T21:47:59.333

1

I would be content if this computation were done over this list in OEIS

Quick & dirty Maple code below checks that only $n=39$ in the list above corresponds to a prime power $\frac{\sigma({39}^2)}{39} = 61$.

vals:=[seq(v[2],v in readdata("b232354.txt",[integer,integer]))]:
for n in vals do:
  m:=NumberTheory:-sigma(n^2)/n:
  ifs:=ifactors(m)[2]:
  comps:=nops(ifs):
  if comps=1 then:
    print(n,ifs[1][1],ifs[1][2]):
  fi:
od:

Output:

39, 61, 1

Though I don't think this helps to conclude anything (that's why this was originally posted just as a comment), because even if the list covers all $n\leq 677004885711$, it is still a list with just $200$ elements. $\sigma(n^2)/n$ being and integer is strict by itself, so adding another condition (it being a prime power) can simply lead to larger example (I am sure someone here could provide a heuristic for this). I don't think you can move on without understanding or at least restricting the $n$ for which $\sigma(n^2)/n$ is an integer.

edited Feb 11 '22 at 21:47

answered Feb 11 '22 at 21:38

Sil

16,612

Thank you for your time and attention, and your answer. I actually predict that $61$ does divide an odd perfect number, which led me to this inquiry. (I think the proof can be done via a factor chain implementation.) That $61$ does divide an odd perfect number $q^k n^2$ with special prime $q$ (yes, using the same notation as in my original post here) will then result to a contradiction, @Sil. – Jose Arnaldo Bebita Dris Feb 11 '22 at 23:53
My apologies for being stubborn, @Sil, but could you check (via Maple) what is the proportion of numbers $n$ satisfying $n \mid \sigma(n^2)$ and ${61} \mid n$, provided $n$ is in this list? I conjecture that this percentage is at least $90 %$. (I do not have Maple installed on my machine, and I do not want to install a pirated copy.) Thanks! – Jose Arnaldo Bebita Dris Feb 12 '22 at 01:30
1

@ARNIEBEBITA-DRIS All of them satisfy $n \mid \sigma(n^2)$ by definition, and $67$ of them satisfy $61 \mid n$. – Sil Feb 12 '22 at 08:52
1

Thanks for checking, @Sil! – Jose Arnaldo Bebita Dris Feb 12 '22 at 15:03

Does there exist a nontrivial prime power $q^k$ such that $\sigma(n^2)/n = q^k$ for some $n$?

1 Answers1