This question was posted in MSE in early August 2020. It did garner several upvotes, but did not receive any responses. I have therefore cross-posted it here, hoping that it gets answered.

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

Here is my question:

Does the equation $sigma(sigma(x^2))=2xsigma(x)$ 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 $sigma(sigma(x^2)) = 2xsigma(x)$ does not have any odd solutions.

Alas, I have no proof.

**SOME PARTIAL RESULTS**

Per Will Jagy’s answer (and a subsequent comment by Erick Wong) to the following MSE question, we have the *conjectured* (sharp?) upper bound

$$frac{sigma(x^2)}{xsigma(x)} leq prod_{rho}{frac{{rho}^2 + rho + 1}{{rho}^2 + rho}} = frac{zeta(2)}{zeta(3)}.$$

*Claim:* $sigma(x^2) neq p^r$ if prime $p geq 5$.

Suppose to the contrary that the equation $sigma(sigma(x^2))=2xsigma(x)$ has an odd solution, and that $sigma(x^2) = p^r$ for some prime $p geq 5$.

Then we have

$$frac{sigma(p^r)}{p^r}=frac{sigma(sigma(x^2))}{sigma(x^2)}=frac{2xsigma(x)}{sigma(x^2)} geq frac{2zeta(3)}{zeta(2)} approx 1.4615259388.$$

But we know that the *abundancy index* $sigma(p^r)/p^r$ is bounded above by

$$frac{sigma(p^r)}{p^r} < frac{p}{p – 1} leq frac{5}{4}.$$

This is a contradiction.

*Claim:* $sigma(x^2) neq p^r q^s$ if primes $p geq 5$ and $q geq 7$.

Suppose to the contrary that the equation $sigma(sigma(x^2))=2xsigma(x)$ has an odd solution, and that $sigma(x^2) = p^r q^s$ for some primes $p geq 5$ and $q geq 7$.

Then we have

$$frac{sigma(p^r)}{p^r}cdotfrac{sigma(q^s)}{q^s}=frac{sigma(sigma(x^2))}{sigma(x^2)}=frac{2xsigma(x)}{sigma(x^2)} geq frac{2zeta(3)}{zeta(2)} approx 1.4615259388.$$

But we know that the *product of abundancy indices* $(sigma(p^r)/p^r)(sigma(q^s)/q^s)$ is bounded above by

$$frac{sigma(p^r)}{p^r}cdotfrac{sigma(q^s)}{q^s} < frac{p}{p – 1}cdotfrac{q}{q – 1} leq frac{5}{4}cdotfrac{7}{6} = frac{35}{24} = 1.458overline{333}.$$

This is a contradiction.