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(The following are taken from this preprint by Antalan and Dris.)

Antalan and Tagle showed that an even almost perfect number $n \neq 2^t$ must necessarily have the form $2^r b^2$ where $r \geq 1$, $\gcd(2,b)=1$, and $b$ is an odd composite.

FIRST CLAIM: $2^r b^2$ is not a square.

PROOF OF FIRST CLAIM: This trivially follows from $\gcd(2,b)=1$ and the fact that $b$ is composite.

Edit: (October 19, 2022 - 5:17 PM Manila time) - This proof for Claim #1 is flawed, please see the rebuttal by Jaap Scherphuis in the comments.

SECOND CLAIM: $2^r b^2$ is not squarefree.

PROOF OF SECOND CLAIM: This also trivially follows from $\gcd(2,b)=1$ and the fact that $b$ is composite.

THIRD CLAIM: $2^r b^2$ has a unique representation as the product of its squarefree part and square part.

PROOF OF THIRD CLAIM: This follows from the fact that $2^r b^2$ is neither a square nor squarefree.

Since $\gcd(2,b)=1$, then the square part of $2^r b^2$ is $b^2$, and the squarefree part of $2^r b^2$ is $2^r$. This would imply that $r \leq 1$. But we know that $r \geq 1$.

Therefore, it follows that $r = 1$.

Furthermore, we then know that $r = 1$ holds if and only if $2^r b^2 \equiv 2 \pmod 3$.

Here is our question:

Is our proof for $r = 1$ logically correct? If not, how can it be mended so as to produce a valid argument?

Jose Arnaldo Bebita Dris
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    Is there an analogue condition for possible positive integers $n$ with $\sigma(n)=2n+1$ ? As far as I know, no such number is known. – Peter Oct 18 '22 at 13:24
  • @Peter: Off-hand, I do not know of an analogue condition for possible positive integers with $\sigma(n) = 2n + 1$. As to their existence, I also think that it is currently an open problem. – Jose Arnaldo Bebita Dris Oct 18 '22 at 14:38
  • @Peter: Found it! This MSE question (and the corresponding answer) proves that, if $\sigma(n) = 2n + 1$, then $n$ is an odd (perfect) square. ("Perfect" in the sense of being a "perfect power", and not in the sense of being a "perfect number".) – Jose Arnaldo Bebita Dris Oct 18 '22 at 14:41
  • $2^rb^2$ will be square whenever $r$ is even, but your proof for claim 1 says nothing about $r$ at all. – Jaap Scherphuis Oct 19 '22 at 09:02
  • Also, doesn't every number have a unique representation as a product of a square and a square-free part, just by unique factorization? – Jaap Scherphuis Oct 19 '22 at 09:06
  • @JaapScherphuis: I was thinking along the lines of "Since $\gcd(2,b)=1$, then $2^r \neq b^2$, which means that $2^r b^2$ cannot be a square". Looking at your rebuttal, I can see that my original argument for Claim 1 is flawed. – Jose Arnaldo Bebita Dris Oct 19 '22 at 09:11
  • @JaapScherphuis: Yes indeed, every number has a unique representation as a product of a square part and a squarefree part, although I am unfamiliar with a proof of this statement that uses unique factorization. – Jose Arnaldo Bebita Dris Oct 19 '22 at 09:12
  • In that case, I don't understand what you are proving in claim 3. Or were you trying to say that $2^r$ is necessarily the square-free part? That isn't true. If the number is not square then $r$ is odd, and the square-free part is $2$ because $2^r$ is divisible by the square $2^{r-1}$. – Jaap Scherphuis Oct 19 '22 at 09:19
  • @JaapScherphuis: Indeed, your point in your last comment is the entire crux of my answer below. (That is, my original proof argument for $r=1$ cannot be right since it would produce an equally dubious proof for $k=1$, where $q^k n^2$ is an odd perfect number with special prime $q$.) – Jose Arnaldo Bebita Dris Oct 19 '22 at 09:27

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This is not a direct answer to the original question, just some remarks that would be too long to fit in the comments.

Blimey! The argument so presented for $r=1$ "cannot be correct". Here is why:

Consider the related problem of determining whether $k=1$ holds, where $q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

(Note that $q^k \neq n^2$ since $\gcd(q,n)=1$. This means that, since $q \geq 5$ and $k \equiv 1 \pmod 4$, then $q^k n^2$ is not a square. Evidently, $q^k n^2$ is not squarefree.)

Since $\gcd(q,n)=\gcd(q^k,n^2)=1$, then the square part of $q^k n^2$ is $n^2$, and the squarefree part of $q^k n^2$ is $q^k$. This would imply that $k \leq 1$. But we know that $k \geq 1$, since $k \equiv 1 \pmod 4$ holds.

Therefore, it follows that $k=1$.

Jose Arnaldo Bebita Dris
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