Does $k=9018009$ have a friend?

Question

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

Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$. For example, $\sigma(6)=1+2+3+6=12$ and $\sigma(28)=1+2+4+7+14+28=56$.

Denote the abundancy index $I$ of $x$ by $$I(x)=\frac{\sigma(x)}{x}.$$

If a positive integer $y$ is one of at least two solutions of $$I(y)=\frac{a}{b}$$ for a given rational number $a/b$, then $y$ is called a friendly number.

Here is a formal definition:

DEFINITION. Let $x$ and $y$ be distinct positive integers. If $x$ and $y$ satisfy the equation $I(x)=I(y)$ then $(x,y)$ is called a friendly pair. Each member of the pair is called a friendly number. (In other words, $x$ is a friend of $y$, and $y$ is also a friend of $x$.) A number which is not friendly is called a solitary number.

Here is my question:

Does $k=9018009$ have a friend?

Answer 1

(Note: This was copied and pasted from an answer (now deleted) to this MO question.)

MOTIVATION

As mentioned in my comment last September $12$, $2016$ to the linked MSE question, $$9018009 = {3003}^2$$ is a factor of the Descartes spoof $198585576189$.

Note that we can write a spoof as $$n = km$$ where $k > 1$, $m > 1$, $m \nmid k$ such that $$\sigma(k)\cdot\left(m + 1\right) = 2n = 2km,$$ where $k$ is an odd square and $m$ is the quasi-Euler prime.

(Note that $22021 = m < k = 9018009$ for the Descartes spoof $198585576189$.)

Now, we can rewrite $$\sigma(k)\cdot\left(m + 1\right) = 2n = 2km$$ as $$\sigma(k) = {m}\cdot\left(2k - \sigma(k)\right),$$ $$\dfrac{\sigma(k)}{m} = 2k - \sigma(k),$$ and $$2k = \left(m + 1\right)\cdot\left(2k - \sigma(k)\right),$$ so that $$\gcd\left(k, \sigma(k)\right) = \gcd\bigg((1/2)\cdot\left(m + 1\right)\cdot\left(2k - \sigma(k)\right), {m}\cdot\left(2k - \sigma(k)\right)\bigg) = 2k - \sigma(k).$$

We therefore have the following lemma:

LEMMA Let $n = km > 1$ be a spoof. Then $$\gcd\left(k, \sigma(k)\right) = 2k - \sigma(k) = \dfrac{\sigma(k)}{m}.$$

The following theorem follows from this lemma:

THEOREM Let $n = km > 1$ be a Descartes spoof. Then $k > 1$ is almost perfect and solitary if and only if $k < m$.

PROOF The details of the proof are in pages $6$ to $7$ of this preprint.

Finally, we have the following conjecture:

CONJECTURE ${3003}^2 = {3^2}\cdot{7^2}\cdot{{11}^2}\cdot{{13}^2} = 9018009$ is not solitary.

Hence, my original inquiry in this MO question.

CONTEXT

My approach to spoofs is similar to that for odd perfect numbers (as initiated in this M. Sc. thesis, pages 98 to 118). As Nielsen has pointed out to me, "the only [case] when results on [spoofs] do not carry over to odd perfect numbers (and vice-versa?) is when the results depend on previous computational results (and assumptions?) on spoofs [reciprocally, odd perfect numbers?]". Indeed, the inequality $m^1 < k$ for the lone Descartes spoof "corresponds" to the inequality $q^k < n^2$ (see this paper) where $q^k n^2$ is an odd perfect number, if we "pretend that $m$ is prime".