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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 repunit, or otherwise? Is there a research work out there that tackles this particular question?

Thanks!

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    $\begingroup$ Is there any reason to expect a result for this special form? $\endgroup$ – Charles Mar 13 '15 at 15:53
  • $\begingroup$ @Charles, well for one, repunit primes are rare, and we haven't found a single odd perfect number (OPN) yet. So maybe (just maybe), the Euler prime of an OPN might be a repunit prime? Your thoughts? $\endgroup$ – Jose Arnaldo Bebita-Dris Mar 13 '15 at 15:57
  • $\begingroup$ Posting a follow-up to this question now. $\endgroup$ – Jose Arnaldo Bebita-Dris Mar 13 '15 at 16:13
  • $\begingroup$ Here is the second question, which I believe is more appropriate for this given problem. $\endgroup$ – Jose Arnaldo Bebita-Dris Mar 13 '15 at 16:17
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NO, the Euler prime $q$ of an odd perfect number $N = {q^k}{n^2}$ is not a repunit, since repunit primes $p > 1$ satisfy $p \equiv 3 \pmod 4$, while it is known that the Euler prime $q$ satisfies $q \equiv 1 \pmod 4$.

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