Let $I(a)=\sigma(a)/a$ denote the abundancy index of the positive integer $a$, where $\sigma(a)=\sigma_1(a)$ is the classical sum of divisors of $a$.
If $I(a)=b$ where $b$ is an integer, then $a$ is called a $b$-multiperfect number.
Here is my question:
Is this an example of a vacuous implication?
If $x$ and $y$ satisfy $\gcd(x,y)=1$, and $I(x)=I(y)=r$, then $x$ and $y$ are both $r$-multiperfect numbers.
Of course, naively considering the condition $\gcd(x,y)=1$ together with $I(x)=I(y)=r$ does imply that $x$ and $y$ are both $r$-multiperfect numbers, since $$r=I(x)=I(y) \iff y\sigma(x)=x\sigma(y)$$ would mean that $$(\gcd(x,y)=1) \land (r=I(x)=I(y)) \implies (x \mid \sigma(x)) \land (y \mid \sigma(y))$$ so that $r$ is an integer.
However, in the case of $r=2$ (for the classical even perfect numbers $M = 2^{p-1}(2^p - 1)$ where $2^p - 1$ is the Mersenne prime of $M$), we do know that $\gcd(M_i, M_j) \geq 2 > 1$ for distinct even perfect numbers $M_i$ and $M_j$.
I currently do not know what happens when $x$ and $y$ are of opposite parity. Hence, this question.
