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(Note: This question is an improved version of and has been cross-posted from this MSE post.)

Let $\sigma(x)$ denote the sum of the divisors of $x$. If $\sigma(x) = 2x - 1$, then we call $x$ an almost perfect number.

If $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r)$ with $r \geq 1$, then $M = {2^r}{b^2}$ is an even almost perfect number.

Now, if $\sigma(2^r)$ is prime, then ${2^r}{\sigma(2^r)}$ is an even perfect number.

In an answer to the linked MSE question, Giovanni Resta asserts that:

If $b = 3^k$ for $k > 1$ then $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q = 2$. Up to ${10}^8$ there are no other values of $b$ that make $q$ prime.

However, by work of Antalan, it is known that $3 \nmid b$, if $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r)$.

Hence, I cannot help but ask:

(1) Is it possible to prove that if $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is prime, then necessarily $b = 3^k$ with $k > 1$? If it is not possible by current mathematical methods, what is the main obstruction?

(2) If $b$ is an odd composite and $r \geq 1$, is it possible to prove that $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} \neq \sigma(2^r)$? If it is not possible by current mathematical methods, what is the main obstruction?

(Note: This question is an improved version of and has been cross-posted from this MSE post.)

Let $\sigma(x)$ denote the sum of the divisors of $x$. If $\sigma(x) = 2x - 1$, then we call $x$ an almost perfect number.

If $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r)$ with $r \geq 1$, then $M = {2^r}{b^2}$ is an even almost perfect number.

Now, if $\sigma(2^r)$ is prime, then ${2^r}{\sigma(2^r)}$ is an even perfect number.

In an answer to the linked MSE question, Giovanni Resta asserts that:

If $b = 3^k$ for $k > 1$ then $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q = 2$. Up to ${10}^8$ there are no other values of $b$ that make $q$ prime.

However, by work of Antalan, it is known that $3 \nmid b$, if $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r)$ with $r \geq 1$.

Hence, I cannot help but ask:

(1) Is it possible to prove that if $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is prime, then necessarily $b = 3^k$ with $k > 1$? If it is not possible by current mathematical methods, what is the main obstruction?

(2) If $b$ is an odd composite and $r \geq 1$, is it possible to prove that $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} \neq \sigma(2^r)$? If it is not possible by current mathematical methods, what is the main obstruction?

(Note: This question is an improved version of and has been cross-posted from this MSE post.)

Let $\sigma(x)$ denote the sum of the divisors of $x$. If $\sigma(x) = 2x - 1$, then we call $x$ an almost perfect number.

If $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r)$ with $r \geq 1$, then $M = {2^r}{b^2}$ is an even almost perfect number.

Now, if $\sigma(2^r)$ is prime, then ${2^r}{\sigma(2^r)}$ is an even perfect number.

In an answer to the linked MSE question, Giovanni Resta asserts that:

If $b = 3^k$ for $k > 1$ then $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q = 2$. Up to ${10}^8$ there are no other values of $b$ that make $q$ prime.

However, by work of Antalan, it is known that $3 \nmid b$.

Hence, I cannot help but ask:

(1) Is it possible to prove that if $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is prime, then necessarily $b = 3^k$ with $k > 1$? If it is not possible by current mathematical methods, what is the main obstruction?

(2) If $b$ is an odd composite and $r \geq 1$, is it possible to prove that $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} \neq \sigma(2^r)$? If it is not possible by current mathematical methods, what is the main obstruction?

(Note: This question is an improved version of and has been cross-posted from this MSE post.)

Let $\sigma(x)$ denote the sum of the divisors of $x$. If $\sigma(x) = 2x - 1$, then we call $x$ an almost perfect number.

If $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r)$ with $r \geq 1$, then $M = {2^r}{b^2}$ is an even almost perfect number.

Now, if $\sigma(2^r)$ is prime, then ${2^r}{\sigma(2^r)}$ is an even perfect number.

In an answer to the linked MSE question, Giovanni Resta asserts that:

If $b = 3^k$ for $k > 1$ then $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q = 2$. Up to ${10}^8$ there are no other values of $b$ that make $q$ prime.

However, by work of Antalan, it is known that $3 \nmid b$, if $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r)$.

Hence, I cannot help but ask:

(1) Is it possible to prove that if $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is prime, then necessarily $b = 3^k$ with $k > 1$? If it is not possible by current mathematical methods, what is the main obstruction?

(2) If $b$ is an odd composite and $r \geq 1$, is it possible to prove that $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} \neq \sigma(2^r)$? If it is not possible by current mathematical methods, what is the main obstruction?

(Note: This question is an improved version of and has been cross-posted from this MSE post.)

Let $\sigma(x)$ denote the sum of the divisors of $x$. If $\sigma(x) = 2x - 1$, then we call $x$ an almost perfect number.

If $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r)$ with $r \geq 1$, then $M = {2^r}{b^2}$ is an even almost perfect number.

Now, if $\sigma(2^r)$ is prime, then ${2^r}{\sigma(2^r)}$ is an even perfect number.

In an answer to the linked MSE question, Giovanni Resta asserts that:

If $b = 3^k$ for $k > 1$ then $q = 2$. Up to ${10}^8$ there are no other values of $b$ that make $q$ prime.

However, by work of Antalan, it is known that $3 \nmid b$.

Hence, I cannot help but ask:

(1) Is it possible to prove that if $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is prime, then necessarily $b = 3^k$ with $k > 1$? If it is not possible by current mathematical methods, what is the main obstruction?

(2) If $b$ is an odd composite and $r \geq 1$, is it possible to prove that $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} \neq \sigma(2^r)$? If it is not possible by current mathematical methods, what is the main obstruction?

(Note: This question is an improved version of and has been cross-posted from this MSE post.)

Let $\sigma(x)$ denote the sum of the divisors of $x$. If $\sigma(x) = 2x - 1$, then we call $x$ an almost perfect number.

If $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r)$ with $r \geq 1$, then $M = {2^r}{b^2}$ is an even almost perfect number.

Now, if $\sigma(2^r)$ is prime, then ${2^r}{\sigma(2^r)}$ is an even perfect number.

In an answer to the linked MSE question, Giovanni Resta asserts that:

If $b = 3^k$ for $k > 1$ then $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q = 2$. Up to ${10}^8$ there are no other values of $b$ that make $q$ prime.

However, by work of Antalan, it is known that $3 \nmid b$.

Hence, I cannot help but ask:

(1) Is it possible to prove that if $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is prime, then necessarily $b = 3^k$ with $k > 1$? If it is not possible by current mathematical methods, what is the main obstruction?

(2) If $b$ is an odd composite and $r \geq 1$, is it possible to prove that $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} \neq \sigma(2^r)$? If it is not possible by current mathematical methods, what is the main obstruction?