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(Note: This question has been cross-posted to MO.)

Let $\sigma(x)$ denote the classical sum of divisors of the positive integer $x$.

Here is my question:

Does the equation $\sigma(\sigma(x^2))=2x\sigma(x)$ have any odd solutions?

MY ATTEMPT

I tried searching for solutions to the equation in Sage Cell Server, in the range $1 < x \leq {10}^6$, here is the Pari-GP code:

for(x=1, 1000000, if(sigma(sigma(x^2))==2*x*sigma(x),print(x,factor(x))))

Here are the results:

9516[2, 2; 3, 1; 13, 1; 61, 1]
380640[2, 5; 3, 1; 5, 1; 13, 1; 61, 1]

Note that both results obtained $x_1 = 9516$ and $x_2 = 380640$ are even.

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

CONJECTURE

The equation $\sigma(\sigma(x^2)) = 2x\sigma(x)$ does not have any odd solutions.

Alas, I have no proof.

Note that, if $x=p$ is prime, then $$p^2 + p + 1 < \sigma(p^2 + p + 1)=\sigma(\sigma(p^2)) = 2p(p+1) = 2(p^2 + p) \implies 1 < p^2 + p,$$ where the last inequality is, of course, trivial.

If $x = q^k$ is a prime power, then $$\frac{q^{2k+1}-1}{q-1} < \sigma\bigg(\frac{q^{2k+1}-1}{q-1}\bigg) = \sigma(\sigma(q^{2k})) = 2q^k \sigma(q^k) = \frac{2q^k \bigg(q^{k+1} - 1\bigg)}{q - 1} = \frac{2q^{2k+1} - 2q^k}{q - 1} \implies 2q^k < q^{2k+1}$$ $$\implies q^{k+1} > 2,$$ where again the last inequality is trivial.

This is where I get stuck.

Jose Arnaldo Bebita Dris
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    No further solution upto $\ x=3\cdot 10^7\ $. – Peter Aug 04 '20 at 08:47
  • Thank you for checking, @Peter. – Jose Arnaldo Bebita Dris Aug 04 '20 at 08:48
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    I'm not sure if this is helpful, but if $x=3\cdot 13\cdot 61\cdot m$ where $m$ is odd such that $\gcd(m,3\cdot 13\cdot 61)=\gcd(\sigma(m^2),8999757)=1$, then we get $7\sigma(\sigma(m^2))=8m\sigma(m)$. Moreover, if $m=7s$ where $s$ is odd such that $\gcd(s,7)=\gcd(\sigma(s^2),57)=1$, then we get $5\sigma(\sigma(s^2))=4s\sigma(s)$. Moreover, if $s=5t$ where $t$ is odd such that $\gcd(t,5)=\gcd(\sigma(t^2),31)=1$, then we get $4\sigma(\sigma(t^2))=3t\sigma(t)$. – mathlove Aug 15 '20 at 17:24
  • @mathlove: It would perhaps be more fruitful to consider the cases when (1) $3 \parallel x$, (2) $3 \not\parallel x \text{ but } 3 \mid x$, and (3) $3 \nmid x$, although honestly speaking, I do not really know how we would deal with Case (3). – Jose Arnaldo Bebita Dris Aug 17 '20 at 01:09
  • @mathlove: Kindly write out your comment from August 15, 2020 as an actual answer to this question, so that I could go ahead and upvote it! My apologies if I missed your comment. – Jose Arnaldo Bebita Dris May 10 '23 at 23:01

1 Answers1

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On OP's request, I am converting my comment into an answer.

I'm not sure if this is helpful, but if $x=3\cdot 13\cdot 61\cdot m$ where $m$ is odd such that $\gcd(m,3\cdot 13\cdot 61)=\gcd(\sigma(m^2),8999757)=1$, then we get $7\sigma(\sigma(m^2))=8m\sigma(m)$.

Moreover, if $m=7s$ where $s$ is odd such that $\gcd(s,7)=\gcd(\sigma(s^2),57)=1$, then we get $5\sigma(\sigma(s^2))=4s\sigma(s)$.

Moreover, if $s=5t$ where $t$ is odd such that $\gcd(t,5)=\gcd(\sigma(t^2),31)=1$, then we get $4\sigma(\sigma(t^2))=3t\sigma(t)$.

Jose Arnaldo Bebita Dris
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mathlove
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