# nt.number theory – Does the equation \$sigma(sigma(x^2))=2xsigma(x)\$ have any odd solutions? 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
$$frac{sigma(p^r)}{p^r} < frac{p}{p – 1} leq frac{5}{4}.$$

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