Let $$\sigma(x) = \sum_{d \mid x}{d}$$ denote the sum of divisors of $x \in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers or positive integers.
Recall that a Descartes number is an odd number $n = km$, with $1 < k$, $1 < m$, satisfying $$\sigma(k)(m+1)=2km.$$ ($m$ is called the quasi-Euler prime of $n$.) Note that we define $\sigma(m) := m + 1$ even when $m$ is composite (that is, we pretend that $m$ is prime).
Notice that the lone Descartes number that is known is $$\mathscr{D} = k'm' = {{3003}^2}\cdot{22021}.$$
In particular, note that:
$k$ is a square.
$\sigma(k)/m = 2k - \sigma(k)$
$m \equiv 1 \pmod 4$
Note that, for the lone Descartes number $\mathscr{D}$ that we know, $$\sqrt{k'} = 3003,$$ which happens to be a palindrome.
Furthermore, since $$\sqrt{k'} = 3 \cdot 7 \cdot {11} \cdot {13},$$ then $\sqrt{k'}$ is squarefree.
Here is my inquiry:
QUESTION: Given an arbitrary Descartes number $n = km > 1$ with quasi-Euler prime $m$, must it necessarily be the case that $\sqrt{k}$ is a squarefree palindrome?
Alas, this is where I get stuck, as we currently do not have another example of a Descartes number.