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!
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 $q \equiv 1 \pmod 4$.