Theorem: If an odd perfect number exists, then it has an odd number of odd proper divisors.
$(1)$ Assume that an odd perfect number exists. Call it $n$.
$(2)$ $2$ is not a divisor because $n$ is odd.
$(3)$ This means that all divisors are odd.
$(4)$ If $n$ has an even number of odd proper divisors, then $n$ is even because the sum of an even number of odd numbers is even. This contradicts our original assumption that $n$ is odd.
$(5)$ So an odd perfect number, if it exists, must have an odd number of odd proper divisors.
This would also satisfy the fact that a perfect number is equal to half the sum of all its divisors because if $n$ is odd and the sum of its proper divisors is odd, then the sum of $n$ and all its divisors would be even and divisible by 2.
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$\begingroup$Your proof is correct, but you could expand it a bit more. When I first read it I was confused with point $(4)$ until I realized you were using the fact that the sum of the proper divisors of a perfect number is that perfect number. It would be good to explain this point.
$\endgroup$ 2 $\begingroup$No, it is not correct. The reason: a perfect number is not a number equal to the sum of its divisors, but a number equal to the sum of its divisors, itself excluded.
So if $n$ is a perfect number, the sum of its divisors is $2n$, which is even, of course.
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