# Is the Euler prime of an odd perfect number a repunit, or otherwise?

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!

• Is there any reason to expect a result for this special form? – Charles Mar 13 '15 at 15:53
• @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? – Jose Arnaldo Bebita-Dris Mar 13 '15 at 15:57
• Posting a follow-up to this question now. – Jose Arnaldo Bebita-Dris Mar 13 '15 at 16:13
• Here is the second question, which I believe is more appropriate for this given problem. – Jose Arnaldo Bebita-Dris Mar 13 '15 at 16:17

## 1 Answer

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 $q \equiv 1 \pmod 4$.