Find the fallacy in the following argument:
Let $N$ be a perfect number, then $N = 2^{p−1} (2^p− 1)$ where $2^p- 1$ is prime. Hence every perfect number is even.
The flaw is at the beginning. Theorem (4.30) requires that $N$ be a perfect even number, before we can conclude it is of the form $2^{p-1}(2^p-1)$ where $(2^p-1)$ is prime.
That is:
Incorrect: Let $N$ be a perfect number, then $N = 2^{p−1} (2^p− 1)$ where $2^p- 1$ is prime.
Correct: Let $N$ be an even perfect number, then $N = 2^{p−1} (2^p− 1)$ where $2^p- 1$ is prime.