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 $1$

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

*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.

(

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

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 $p \geq 3$.

This results to the contradiction

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

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

$$N = {p^{2k}}{q^{2l}} \geq {p^2}{q^2} = (pq)^2 \geq ({3}\cdot{5})^2 = {15}^2 = 225.$$

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

$$I(N) = \frac{\sigma(N)}{N} = \frac{\sigma({p^{2k}}{q^{2l}})}{{p^{2k}}{q^{2l}}} = \frac{\left(p^{2k+1} - 1\right)\left(q^{2l + 1} - 1\right)}{{p^{2k}}{q^{2l}}\left(p - 1\right)\left(q - 1\right)}$$

(

*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 $p \geq 3$.

This results to the contradiction

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

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 $\omega(N) \geq 3$.

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:

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

$$\frac{2N}{N + 1} < I(N) = \frac{\sigma(N)}{N} < \prod_{i=1}^{M}{\frac{p_i}{p_i - 1}}$$

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.