# On odd perfect numbers $N$ given in the Eulerian form $N = {q^k}{n^2}$

Note: This question was cross-posted from MO.

Preamble: I apologize in advance if this particular MSE post would appear to be a bit of a polymath approach, I just had to put down all the details to present my argument for this particular math problem.

A positive integer $N$ is said to be perfect if $\sigma(N)=2N$, where $\sigma(x)$ is the sum of the divisors of $x$.

An odd perfect number $N$ is said to be given in Eulerian form if $N = {q^k}{n^2}$, where $q$ is prime, $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Since $\gcd(q,n)=1$ and prime powers are deficient, we have $q \neq n$ and $q^k \neq n$.

In an earlier version of the paper The Abundancy Index of Divisors of Odd Perfect Numbers (see here), it was conjectured that the biconditional

$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$

is true. (Note that the proof of the inequation

$$\frac{\sigma(q^k)}{n} \neq \frac{\sigma(n)}{q^k}$$

is trivial.)

Recently, an attempt to prove the said biconditional appears to have been completed in this preprint.

I present here the highlights of the said proof:

One direction of the biconditional is trivial:

$$q^k < n \Longrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}.$$

This is proved by noting that $I(q^k) < \sqrt[3]{2} < I(n)$ (where $I(x) = \sigma(x)/x$ is the abundancy index of $x$), from which the following chain of implications follow:

$$q^k < n \Longrightarrow \sigma(q^k) < \sigma(n)$$ $$q^k < n \Longrightarrow \frac{1}{n} < \frac{1}{q^k}$$ $$\{\sigma(q^k) < \sigma(n)\} \land \{\frac{1}{n} < \frac{1}{q^k}\} \Longrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$

Therefore:

$$q^k < n \Longrightarrow \sigma(q^k) < \sigma(n) \Longrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}.$$

For the other direction:

The implication

$$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k} \Longrightarrow \sigma(q^k) < \sigma(n)$$ can be proved, again by observing that $I(q^k) < \sqrt[3]{2} < I(n)$.

Now, to prove the last implication:

$$\sigma(q^k) < \sigma(n) \Longrightarrow q^k < n$$

we take an indirect approach.

First we show that:

If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form and

$$I(q^k) + I(n) < \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k},$$

then $q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n)$.

Similarly, we can prove that:

If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form and

$$\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} < I(q^k) + I(n),$$

then $q^k < n \Longleftrightarrow \sigma(n) < \sigma(q^k)$.

Observe that

$$I(q^k) + I(n) = \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$

if and only if

$$\sigma(q^k) = \sigma(n).$$

Also, observe that if we assume

$$\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} < I(q^k) + I(n)$$

then the biconditional

$$q^k < n \Longleftrightarrow \sigma(n) < \sigma(q^k)$$

will contradict $I(q^k) < \sqrt[3]{2} < I(n)$.

Therefore, the following inequality must be true:

$$I(q^k) + I(n) \leq \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}.$$

It suffices to consider the case when

$$I(q^k) + I(n) = \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$

which is true if and only if

$$\sigma(q^k) = \sigma(n).$$

This last equation, together with the inequality $I(q^k) < \sqrt[3]{2} < I(n)$, implies that

$$1 = \frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n}$$

from which it follows that $n < q^k$. Thus, $1/q^k < 1/n$, which then gives, together with the equation $\sigma(n) = \sigma(q^k)$, the inequality

$$\frac{\sigma(n)}{q^k} < \frac{\sigma(q^k)}{n}.$$

But this last inequality, together with $I(q^k) < \sqrt[3]{2} < I(n)$, is known to imply

$$n < q^k.$$

Consequently, if

$$I(q^k) + I(n) = \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$

then

$$n < q^k \Longleftrightarrow \frac{\sigma(n)}{q^k} < \frac{\sigma(q^k)}{n}$$

Now, here is the part where I am a bit unsure about logical solidity:

Lastly, note that since

$$\sigma(q^k) = \sigma(n)$$

implies $n < q^k$, the biconditional

$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n)$$

is vacuously true, under this case.

NOW HERE IS MY QUESTION (and this is also the main reason for this MSE post):

I have been told that there is a gap in the last part of this proof. As I myself am having a hard time spotting where that particular error is, would somebody be kind enough as to help by skimming through this argument and then (after glossing over the details) give it either a PASS or a FAIL?

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I do not see why $\sigma(q^k)=\sigma(n)$ would imply $q^k>n$. For example, $$\sigma(3^7)=\sigma(2863)=\sigma(3097)=3280,$$ but $$3^7=2187<2863<3097.$$ – Librecoin Nov 3 '13 at 11:27
On the other hand, $$\sigma(5^3)=\sigma(99)=156$$ and $$5^3=125>99,$$ so $\sigma(q^k)=\sigma(n)$ does not imply any relation between $q^k$ and $n$. – Librecoin Nov 3 '13 at 11:37
Thank you for the "counterexamples", @Tharsis. – Arnie Dris Nov 4 '13 at 14:30
However, in view of the inequality $I(q^k) < 5/4 < \sqrt{8/5} < I(n)$ (see here for a proof), we do have $$1 = \frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n}$$ from which it follows that $\sigma(q^k) = \sigma(n)$ does imply $q^k > n$ -- all of these of course under the assumption that there exists an odd perfect number $N = {q^k}{n^2}$. – Arnie Dris Nov 4 '13 at 14:34
Besides, please do note that for an odd perfect number $N$ given in the Eulerian form $N = {q^k}{n^2}$, the following must be true: (1) $q$ is prime with $q \equiv k \equiv 1 \pmod 4$, (2) $\gcd(q, n) = 1$. Let me know if you have any more questions and / or comments, @Tharsis! Appreciate it... =) – Arnie Dris Nov 4 '13 at 14:39

Recall that, since $\gcd(q,n)=1$, we have $q^k \neq n$.

Note that, from the original post, we have a proof for the biconditional

$$q^k < n \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$

if $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form.

I give a proof for the following biconditional in this answer:

$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n).$$

So we have two cases to consider:

Case 1: $I(q^k) + I(n) < \sigma(q^k)/n + \sigma(n)/q^k$

Under this case, we already have

$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n).$$

Case 2: $I(q^k) + I(n) = \sigma(q^k)/n + \sigma(n)/q^k$

In this case, we have $\sigma(q^k) = \sigma(n)$, which implies, via $I(q^k) < I(n)$, $n < q^k$. Hence, $q^k < n$ and $\sigma(q^k) < \sigma(n)$ are both false. This implies that the biconditional

$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n)$$

is vacuously true, under Case 2. (Equivalently, the biconditional

$$n < q^k \Longleftrightarrow \sigma(n) \leq \sigma(q^k)$$

is true.)

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Update (January 30, 2015)

Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form.

While $q^k < n$ does imply $\sigma(q^k) < \sigma(n)$ (via $I(q^k) < I(n)$), $\sigma(q^k) < \sigma(n)$ by itself DOES NOT imply $$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}.$$ Thus, the implication $$\sigma(q^k) < \sigma(n) \implies \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$ remains unproven.

Similarly, while $$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$ does imply $\sigma(q^k) < \sigma(n)$ (via $I(q^k) < I(n)$), $\sigma(q^k) < \sigma(n)$ by itself DOES NOT imply $q^k < n$.

Thus, the implication $\sigma(q^k) < \sigma(n) \implies q^k < n$ remains unproven.

Salvaging Results

We consider two cases:

Case A: $\sigma(q^k) \neq \sigma(n)$

Under Case A, we have $\sigma(q^k) < \sigma(n) \Longleftrightarrow q^k < n$. Consequently, the biconditional $$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$ holds.

Case B: $\sigma(q^k) = \sigma(n)$

Under Case B, since $I(q^k) < I(n)$, we have $n < q^k$. Since the underlying assumption is $\sigma(q^k) = \sigma(n)$, we get that $$\sigma(n) = \sigma(q^k) \implies \left(n < q^k \implies \frac{\sigma(n)}{q^k} < \frac{\sigma(q^k)}{n}\right).$$

But from $I(q^k) < I(n)$, we know that the following implication must also be true: $$\frac{\sigma(n)}{q^k} < \frac{\sigma(q^k)}{n} \implies n < q^k.$$

Consequently, under Case B, the biconditional $$q^k < n \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$ holds.

THEREFORE, IN GENERAL WE OBTAIN

If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then the biconditional $$q^k < n \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$ holds.

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