Divisibility of odd numbers and its sum of divisors function

Question

Let us denote as $d(n)$ some proper divisor of $n$ such that $n$ is odd.

I found recently the following

Theorem

If $n=p^\alpha*q$, where $p$ and $q$ are prime numbers, and $q=p^\alpha-\frac{p^\alpha-1}{p-1}$, then $$\sum_{d\leq\sqrt{n}}d\left(n\right)\mid n$$

The proof relies on the fact that, as $p^\alpha>q$, then the sum of proper divisors of $n$ which are less than the square root of $n$ is $$\frac{p^\alpha-1}{p-1}+q=p^\alpha$$

I was wondering if

1. Could be this theorem biconditional? That is, could it be proved that $\sum_{d\leq\sqrt{n}}d\left(n\right)\mid n$ only if $n=p^\alpha*q$?
2. If answer to question 1 is negative (this theorem is not biconditional), could this theorem be true also for $n=p^\alpha*q^\beta$ such that $\beta>1$? How to prove it?

Thanks in advance!

Answer 1

Hint: Your formula for $q$ is equal to $$q = p^{\alpha} - \frac{p^{\alpha} - 1}{p - 1} = p^{\alpha} - \bigg(\sigma(p^{\alpha}) - p^{\alpha}\bigg) = 2p^{\alpha} - \sigma(p^{\alpha}) = D(p^{\alpha}),$$ where $D(x)$ is the deficiency of the positive integer $x$, and $\sigma$ is the classical sum-of-divisors function.