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If $N = qn^2$ is an odd perfect number with $\gcd(q,n)=1$, is it possible to have $q + 1 = \sigma(n)$?
The title says it all.
Question
If $N = qn^2$ is an odd perfect number with Euler prime $q$ and $\gcd(q,n)=1$, is it possible to have $q + 1 = \sigma(n)$?
Heuristic
From the Descartes spoof, with ...
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On even almost perfect numbers other than powers of two
(Note: This question is an improved version of and has been cross-posted from this MSE post.)
Let $\sigma(x)$ denote the sum of the divisors of $x$. If $\sigma(x) = 2x - 1$, then we call $x$ an ...