Is the odd part of even almost perfect numbers (other than the powers of two) not almost perfect? Let $\sigma(x)$ denote the sum of the divisors of $x$.  A number $M$ is called almost perfect if $\sigma(M) = 2M - 1$.
If $M$ is an even almost perfect number, then the only known examples for $M$ are $M = 2^k$, where $k \geq 1$.  (Since $\sigma(1) = 2\cdot{1} - 1$, $1$ is an odd almost perfect and is currently the only known such number.)
Antalan and Tagle (in a 2004 preprint titled Revisiting forms of almost perfect numbers) show that, if $N \neq 2^k$ is an even almost perfect number, then $N$ takes the form $N = {2^r}{b^2}$ where $r \geq 1$ and $b$ is an odd composite.  We call $b^2$ the odd part of the even almost perfect number $N$.
Since $N$ is almost perfect, we have
$$(2^{r+1} - 1)\sigma(b^2) = \sigma(2^r)\sigma(b^2) = \sigma({2^r}{b^2}) = \sigma(N) = 2N - 1 = {2^{r+1}}b^2 - 1.$$
So we have
$$\sigma(b^2) = \frac{{2^{r+1}}b^2 - 1}{2^{r+1} - 1} = b^2 + \frac{b^2 - 1}{2^{r+1} - 1}.$$
Now,
$$2b^2 - \sigma(b^2) = b^2 -\frac{b^2 - 1}{2^{r+1} - 1}.$$
If the odd part $b^2$ is also almost perfect, then we have
$$1 = 2b^2 - \sigma(b^2) = b^2 -\frac{b^2 - 1}{2^{r+1} - 1},$$
which, since $b > 1$, gives
$$2^{r+1} - 1 = 1 \Longleftrightarrow r = 0.$$
This contradicts $r \geq 1$.
Consequently, since $b^2$ is deficient, we can write $\sigma(b^2) = 2b^2 - c$, where $c > 1$. (We then call $b^2$ a $c$-deficient number.)
My question is:  Is this proof correct?


Is this proof correct?

I have not found any errors in your proof, so it looks correct to me.
