Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form (i.e., $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n) = 1$). ($\sigma(x)$ gives the sum of the divisors of $x$, and $y$ is perfect if $\sigma(y) = 2y$.)

In this paper, it is shown that the inequality $n < q$ is sufficient for Sorli's conjecture that $k = 1$ to hold.

Somebody pointed out to me that it is obvious that the following implication is true:

$$\sigma(n) \leq q^k \Longrightarrow k = 1$$

Is it indeed so? It's not obvious to me for two reasons:

  1. $$n < q^k \Longrightarrow \left\{n < q \Longleftrightarrow k = 1\right\}$$

  2. $$k > 1 \Longrightarrow q < n$$

That is, in order to prove the supposedly obvious claim, it would suffice to show that the condition

$$q < n < q^k$$

never holds.


1 Answer 1


After getting back to my correspondent, the claim that

$$\sigma(n) \leq q^k \Longrightarrow k = 1,$$

where $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, has been retracted.


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