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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 $$k = 1 \implies q = 5 \implies k = 1.$$

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    $\begingroup$ 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. $\endgroup$ – Pace Nielsen Dec 29 '16 at 21:10
  • $\begingroup$ Thanks for pointing that out, Pace! I have noted your observation in the question. $\endgroup$ – Arnie Bebita-Dris Dec 29 '16 at 21:37
  • $\begingroup$ @PaceNielsen, I dimly recall that Hagis (there may be others, of course) proved some results similar to those in your comment. Notwithstanding, do you happen to know if $q = 5 \implies k = 1$ holds unconditionally? $\endgroup$ – Arnie Bebita-Dris Dec 29 '16 at 21:41
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    $\begingroup$ I think that's a wide-open problem. $\endgroup$ – Pace Nielsen Dec 29 '16 at 22:48
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The equivalence :

$(k=1 \lor q> 5) \iff (q=5 \implies k=1)$

is not correct on its own, except if you have some other information (I don't know the problem, I'm simply evaluating the "logical" points of the proof); that's because $(q=5 \implies k=1)$ is itself equivalent to $(q\neq 5 \lor k=1)$ which is not equivalent to the LHS. What's more, you don't have the $k=1 \implies q=5$ part of your chain of implications (*) (and that's the critical point I think), what you have is (if the above equivalence is correct for some reason)

$k=1 \implies (q=5 \implies k=1)$,

which is a wildly different statement, as for any formula $p$, we have $k=1 \implies (p\implies k=1)$. So unless I'm making a stupid mistake somewhere, no this proof is not "logically sound"

(*) it's important to remember a chain of implications from which you can deduce an equivalence is, formally, a formula like $(A_1 \implies A_2) \land .... \land (A_n \implies A_1)$, which cannot be abbreviated as $A_1 \implies A_2 \implies ... \implies A_n \implies A$, because $\implies$ is not associative

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    $\begingroup$ Also, reading with sufficient care the section containing Lemma 12 gives more assumptions on N than are stated above (in particular all but two of the prime factors are less than 100). We do not have either direction of the general bi-conditional. Gerhard "Sounded Too Good As Truth" Paseman, 2016.12.29. $\endgroup$ – Gerhard Paseman Dec 29 '16 at 21:11

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