Is the Euler prime of an odd perfect number a palindrome (in base $10$), or otherwise?

Question

Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form (i.e., $q$ is prime with $\gcd(q,n)=1$ and $q \equiv k \equiv 1 \pmod 4$). (That is, $2N=\sigma(N)$ where $\sigma$ is the classical sum-of-divisors function.)

Since $\gcd(q^k,\sigma(q^k))=1$, it follows that $q \mid \sigma(n^2)$.

My question is this:

Is the Euler prime $q$ of an odd perfect number a palindrome in base $10$, or otherwise? Is there a research work out there that tackles this particular question?

Thanks!

Answer 1

Reading through the Wikipedia page on palindromic numbers:

G. J. Simmons conjectured there are no palindromes of form $r^s$ for $s > 4$ (and $r > 1$).

So I guess a more appropriate question for the original context of this post would have been:

Is the Euler factor $q^k$ of an odd perfect number $N={q^k}{n^2}$ a palindrome in base $10$, or otherwise ?

Note that the condition $k = 1$ is called the Descartes-Frenicle-Sorli conjecture for odd perfect numbers.

The citation for the Simmons reference is as follows:

G. J. Simmons, Palindromic powers, Journal of Recreational Mathematics, 3 (No. 2, 1970), 93-98