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?


  • 4
    $\begingroup$ Being a palindrome is base dependent (i.e., a number may be a palindrome in base ten (decimal system), but not in some other base). So it seems unlikely that $q$ being or not being a palindrome would have much to do with the perfect number question. Of course if it turns out that there are no odd perfect numbers, then the Euler prime of every odd perfect number would indeed be a palindrome. (Or not a palindrome if you prefer.) $\endgroup$
    – paw88789
    Commented Mar 13, 2015 at 16:23

1 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


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