If $2^r b^2$ is an even almost perfect number that is NOT a power of two, does it follow that $r=1$?

Question

(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.

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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.

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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.

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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.

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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 $2^r b^2 \equiv 2 \pmod 3$.

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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?

Answer 1

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

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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 $k=1$.