## 14.7.16

### Odd Numbers Ruled Out by the Dris Criterion on Almost Perfect Numbers (Using the Abundancy Index)

In this post, we consider odd numbers that are ruled out by the Dris theorem, which gives the following criterion for almost perfect numbers in terms of the abundancy index:

Theorem.  (Dris)  Let $N$ be a positive integer.  Then $\sigma(N) = 2N - 1$ if and only if
$$\frac{2N}{N + 1} \leq I(N) < \frac{2N + 1}{N + 1}.$$
Equality holds if and only if $N = 1$.

First, note that $N_0 = 2^r$ satisfies the inequality in the criterion, as WolframAlpha confirms that
$$\frac{2^{r+1}}{2^r + 1} \leq \frac{2^{r+1} - 1}{2^r} < \frac{2^{r+1} + 1}{2^r + 1}$$
is true for $r \geq 0$.  (Note that $1$ is almost perfect.)

Second, note that if $N$ is odd and $\sigma(N) = 2N - 1$, then $N$ must be a square.

A.  Suppose that $N_1 = p^k$, where $p$ is an odd prime and $k \geq 2$ is an even integer.  We want to show that $N_1$ is not almost perfect.

To this end, assume that
$$\frac{2{p^k}}{p^k + 1} < \frac{p^{k+1} - 1}{{p^k}(p - 1)} < \frac{2{p^k} + 1}{p^k + 1}.$$

WolframAlpha says that this inequality is equivalent to
$$2p^k < -\frac{p^{-k}}{p - 1} + \frac{p^{k+1}}{p - 1} + \frac{p}{p - 1} - \frac{1}{p - 1} < 2{p^k} + 1,$$
which, by a brief inspection, is not true.  (It suffices to check the inequality on the left.)

Since the inequality in the criterion is satisfied when $p = 2$, we now know that $N_1 = p^k$ is not almost perfect, when $p$ is an odd prime with $k \geq 2$.

B.  Suppose that $N_2 = p^k q^l$, where $p$ and $q$ are distinct odd primes, and $k \geq 2$, $l \geq 2$ are even integers.  We want to show that $N_2$ is not almost perfect.

To this end, assume that
$$\frac{2{p^k}{q^l}}{{p^k}{q^l} + 1} < \frac{\left(p^{k+1} - 1\right)\left(q^{l+1} - 1\right)}{{p^k}{q^l}\left(p - 1\right)\left(q - 1\right)} < \frac{2{p^k}{q^l} + 1}{{p^k}{q^l} + 1}.$$

WolframAlpha says that this inequality is equivalent to
$$2{p^k}{q^l} < \frac{{p^{-k}}{q^{-l}}}{\left(p - 1\right)\left(q - 1\right)} + \frac{{p^{k+1}}{q^{l+1}}}{\left(p - 1\right)\left(q - 1\right)} - \frac{q{p^{-k}}}{\left(p - 1\right)\left(q - 1\right)} - \frac{p^{k+1}}{\left(p - 1\right)\left(q - 1\right)}$$
$$-\frac{p{q^{-l}}}{\left(p - 1\right)\left(q - 1\right)} - \frac{q^{l+1}}{\left(p - 1\right)\left(q - 1\right)} + \frac{pq}{\left(p - 1\right)\left(q - 1\right)} + \frac{1}{\left(p - 1\right)\left(q - 1\right)} < 2{p^k} + 1,$$
which, by a brief inspection, is not true.  (Again, it suffices to check the inequality on the left.)

We conclude that $N_2$ is not almost perfect.

(Moving forward, since the inequality on the right appears to be always satisfied, we will only consider the inequality on the left.)

An Aside:  Let $\omega(N)$ denote the number of distinct prime factors of $N$. If $N$ is odd and satisfies $\sigma(N) = 2N - 1$, then an easy way to prove that $\omega(N) \geq 3$ is as follows:

Since $\sigma(N) = 2N - 1$ and $N$ is odd, it follows that $N$ is a square.

Suppose that $\omega(N)=1$.  Then $N$ takes the form $p^{2k}$, where $p$ is an odd prime and $k \geq 1$ is an integer.  Note that $p \geq 3$, from which it follows that $N = p^{2k} \geq p^2 \geq 9$.

We obtain
$$I(N) = \frac{\sigma(N)}{N} = 2 - \frac{1}{N} \geq 2 - \frac{1}{9} = \frac{17}{9} = 1.\overline{888}.$$

But we also have
$$I(N) = \frac{\sigma(N)}{N} = \frac{\sigma(p^{2k})}{p^{2k}} = \frac{p^{2k+1} - 1}{{p^{2k}}\left(p - 1\right)} < \frac{p^{2k+1}}{{p^{2k}}\left(p - 1\right)} = \frac{p}{p - 1} \leq \frac{3}{2} = 1.5,$$
since $p \geq 3$.

$$1.\overline{888} = \frac{17}{9} \leq I(N) < \frac{3}{2} = 1.5.$$

This means that $\omega(N) \geq 2$.  Now, suppose that $\omega(N) = 2$.  Again, as before, $N$ takes the form ${p^{2k}}{q^{2l}}$, where $p$ and $q$ are distinct odd primes (so that we may take $p < q$, without loss of generality) and $k \geq 1$, $l \geq 1$ are integers.  Note that we may take $3 \leq p$ and $5 \leq q$, from which it follows that
$$N = {p^{2k}}{q^{2l}} \geq {p^2}{q^2} = (pq)^2 \geq ({3}\cdot{5})^2 = {15}^2 = 225.$$

As before, we obtain
$$I(N) = \frac{\sigma(N)}{N} = 2 - \frac{1}{N} \geq 2 - \frac{1}{225} = \frac{449}{225} = 1.99\overline{555},$$
since $p \geq 3$ and $q \geq 5$.

But we also have
$$I(N) = \frac{\sigma(N)}{N} = \frac{\sigma({p^{2k}}{q^{2l}})}{{p^{2k}}{q^{2l}}} = \frac{\left(p^{2k+1} - 1\right)\left(q^{2l + 1} - 1\right)}{{p^{2k}}{q^{2l}}\left(p - 1\right)\left(q - 1\right)}$$
$$< \frac{{p^{2k+1}}{q^{2l+1}}}{{p^{2k}}{q^{2l}}\left(p - 1\right)\left(q - 1\right)} = \left(\frac{p}{p - 1}\right)\cdot\left(\frac{q}{q - 1}\right) \leq \left(\frac{3}{2}\right)\cdot\left(\frac{5}{4}\right) = \frac{15}{8} = 1.875$$
since $p \geq 3$ and $q \geq 5$.

This results to the contradiction$$1.99\overline{555} = \frac{449}{225} \leq I(N) < \frac{15}{8} = 1.875.$$

We therefore conclude that $\omega(N) \geq 3$.  QED.

C.  Suppose that $N_3 = p^k q^l r^m$, where $p$, $q$ and $r$ are distinct odd primes, and $k \geq 2$, $l \geq 2$, $m \geq 2$ are even integers.  We want to show that $N_3$ is not almost perfect.

To this end, assume that
$$\frac{2{p^k}{q^l}{r^m}}{{p^k}{q^l}{r^m} + 1} < \frac{\left(p^{k+1} - 1\right)\left(q^{l+1} - 1\right)\left(r^{m+1} - 1\right)}{{p^k}{q^l}{r^m}\left(p - 1\right)\left(q - 1\right)\left(r - 1\right)} < \frac{2{p^k}{q^l}{r^m} + 1}{{p^k}{q^l}{r^m} + 1}.$$

As indicated earlier, we will only be considering the inequality on the left.  This inequality implies
$$2{p^k}{q^l}{r^m} < \left({p^k}{q^l}{r^m} + 1\right)\cdot\left(\frac{p}{p - 1}\cdot\frac{q}{q - 1}\cdot\frac{r}{r - 1}\right),$$
which again, by a brief inspection, is not true.

Generalizing to the case of a fixed $\omega(N)$, we now show that indeed, the following "theorem" holds:

"Theorem".  No odd almost perfect numbers exist apart from $1$.

"Proof".  It suffices to consider the case when we have an odd $N > 1$ and $\omega(N) = M$, where $M$ is fixed.

Assume to the contrary that $N = \prod_{i=1}^{M}{{p_i}^{\alpha_i}}$ is almost perfect.  (Note that the $\alpha_i$'s are all even, but it will not matter in this argument.)  By the Dris criterion, we have
$$\frac{2N}{N + 1} < I(N) = \frac{\sigma(N)}{N} < \prod_{i=1}^{M}{\frac{p_i}{p_i - 1}}$$
so that
$$2{\prod_{i=1}^{M}{{p_i}^{\alpha_i}}} < \left(\left({\prod_{i=1}^{M}{{p_i}^{\alpha_i}}}\right) + 1\right)\cdot\left(\prod_{i=1}^{M}{\frac{p_i}{p_i - 1}}\right).$$

Similar to the reasoning above, by a brief inspection, this inequality cannot be true.

We therefore conclude that $N$ cannot be almost perfect.  Since $M = \omega(N)$ was arbitrary, it follows that there are no odd almost perfect numbers, apart from $1$.

"QED".

Update (Added July 14 2016):  There is a serious flaw in the arguments beginning Part C.  However, by utilizing a result of Kishore that an odd almost perfect number has to have at least six (6) distinct prime factors, it is possible to prove the following result instead:

Theorem.  If $N > 1$ is an odd almost perfect number with $\omega(N)=6$, then $N$ must be divisible by $3$.

Proof.  The details are in this MSE post.