How do I show $1$ is not a trivial odd perfect number?

Question

This question related to my this question in MO ,some comments stated that the integer $1$ is trivial answer for this question ,but here i'm very confused when we say that the sum divisors of $1$ is $1$ it self then easly $1$ is odd perfect number and with who we are sum $1$ , is it with$ 0$ as $1=1+0$ but $0$ is not divisor of any integer number .

Here my question: Is there some who can show me why we can't judge that $1$ is a trivial odd perfect number ?

Thank you for any help

Answer 1

A perfect number is a positive integer that is equal to the sum of its proper positive divisors. That is, we add up all the divisors not including the number itself. Therefore, the sum of all of $1$'s proper divisors is 0, and so it is not perfect.

Edit: We are assuming here, by convention, that the empty sum $=0$

Answer 2

Since the sum-of-divisors function $\sigma$ is weakly multiplicative, then we have $\sigma(1) = 1$.

If $1$ were a classical $2$-perfect number, then $\sigma(1)$ should be equal to $2\cdot1$ (which, of course, is not true).

(It turns out that $1$ is the only odd multiperfect number that is currently known. $1$ is considered to be a $1$-perfect number.)