This question is inspired by this earlier one:
Divisibility of odd numbers and its sum of divisors function
In that question, MSE user Juan Moreno claims to have discovered a proof for the following "Theorem":
"Theorem": Let us denote as $d(n)$ some proper divisor of $n$ such that $n$ is odd. If $n=p^{\alpha} q$, where $p$ and $q$ are prime numbers, and $q=p^{\alpha}-\frac{p^{\alpha}-1}{p-1}$, then $$\sum_{d\leq\sqrt{n}}d\left(n\right)\mid n.$$
Of course, we obtain $$q = p^{\alpha}-\frac{p^{\alpha}-1}{p-1} = p^{\alpha} - \bigg(\sigma(p^{\alpha}) - p^{\alpha}\bigg) = 2p^{\alpha} - \sigma(p^{\alpha}) = D(p^{\alpha}),$$ where $D(x)=2x-\sigma(x)$ is the deficiency of the positive integer $x$. (Here, $\sigma$ is the classical sum-of-divisors function.)
Since $$D(p^{\alpha}) < p^{\alpha}$$ (for prime $p$ and $\alpha \geq 1$), and since $$D(p^{\alpha}) + s(p^{\alpha}) = p^{\alpha},$$ where $s(x)=\sigma(x)-x$ is the sum of aliquot divisors of $x$, is Juan Moreno correct when he claimed that $$\sum_{d\leq\sqrt{n}}d\left(n\right) = D(p^{\alpha}) + s(p^{\alpha}) = p^{\alpha}?$$