(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.