Expressing relationship between Mersenne primes and even perfect numbers as a formula

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I'm trying to understand why the Euclid-Euler theorem implies a one-to-one correspondence between Mersenne primes and perfect numbers.

Theorem 1 (Euclid):

  • If $2^{p}−1$ is a Mersenne prime, then $2^{p−1}⋅(2^{p}−1)$ is perfect.

Theorem 2 (Euler):

  • If $n$ is even and perfect, then there is a Mersenne prime $2^{p}−1$ such that $n = 2^{p−1}(2^{p}−1)$.

Wikipedia and other sources I've come across say that these two theorems together imply a "one-to-one relationship between even perfect numbers and Mersenne primes; each Mersenne prime generates one even perfect number, and vice versa."

But it's not intuitive to me why Theorem 2 is the converse of Theorem 1, since we cannot rule out odd perfects. I would prefer to instead verify this using logic rules, but don't know how to express Theorem 2 as a formula.

Assuming the universe of discourse is the natural numbers, if I let

  • M(x) = "x is a Mersenne prime"
  • P(x) = "x is a perfect number"
  • E(x) = "x is even"

...then I can write Theorem 1 as $\forall p(M(2^{p}−1) \rightarrow P(2^{p−1}(2^{p}−1))$.

But how do I go about expressing Theorem 2 in this way?

I tried the following

$\forall n[(E(n) \land P(n)) \rightarrow \exists p(M(2^{p}−1) \land n = 2^{p−1}(2^{p}−1))]$

...but I doubt this is correct and I'm not sure where to go from here.


EDIT: Someone very helpfully pointed out that the first theorem implies that $2^{p−1}(2^{p}−1)$ is also even (in addition to being perfect), since $p$ is prime and therefore the exponent is at least 1.

Here's my new attempt:

Assume the universe is the natural numbers and let $M(w)$, $P(w)$ and $E(w)$ stand for the following statements:

  • $M(w)$ – “w is Mersenne prime”
  • $P(w)$ – “w is perfect”
  • $E(w)$ – “w is even”

Then I can write Theorem 1 as:

$$∀x[M(2^x - 1) \rightarrow \exists y(y = 2^{x-1}(2^x – 1) ∧ P(y) ∧ E(y))]$$

...and Theorem 2 as:

$$∀x[∃y(y = 2^{x-1}(2^x – 1) ∧ P(y) ∧ E(y)) \rightarrow M(2^x - 1)]$$

But is this equivalent to the statement given above that "If $n$ is even and perfect, then there is a Mersenne prime $2^{p}−1$ such that $n = 2^{p−1}(2^{p}−1)$"? I'm not sure if I've expressed the consequent properly.