Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$.

We want to show that the biconditional $k = 1 \iff q = 5$ holds.

~~It suffices to prove one direction, as the implication $q = 5 \implies k = 1$ was proved by Iannucci (Lemma 12, page 873).~~ Per a comment from Pace: "Jose, you are misinterpreting the result from Iannucci. His number $N$ is restricted to be an odd perfect number which additionally has all but at most two of its prime divisors less than $100$. He goes on to prove that there are no such numbers, but along the way, in the lemma you cite, he shows that such numbers have restricted form."

To this end, we show first that $$k = 1 \implies I(n^2) \leq 2 - \frac{5}{3q}.$$ The proof is easy. We refer the reader to this preprint.

Next, we observe that
$$I(n^2) = 2 - \frac{5}{3q}$$
implies that
$$k = 1 \land q = 5,$$
(The assertion that follows only holds when $k=1$.)
~~while
$$I(n^2) < 2 - \frac{5}{3q}$$
implies $q > 5$.~~

In particular, we have $$I(n^2) \leq 2 - \frac{5}{3q} \implies \bigg(\left(k=1\right) \lor \left(q>5\right)\bigg).$$

But we know that $$\bigg(\left(k=1\right) \lor \left(q>5\right)\bigg) \iff \bigg(q = 5 \implies k = 1\bigg).$$

Consequently, we have the chain of implications $$q = 5 \implies k = 1 \implies q = 5 \implies k = 1,$$ and we are done.

Now here is my question:

Is this "

proof" logically sound?

**Update (December 30 2016)**

As in the given answer, we could only have the implication $$k = 1 \implies \bigg(q = 5 \implies k = 1\bigg),$$ which I agree is not the same as $$k = 1 \implies q = 5 \implies k = 1.$$

not squarefree, if $q^k n^2$ is an odd perfect number with special prime $q$. (I invite you to check out this answer to a closely related MO question.) $\endgroup$