Let $\sigma(x)$ be the sum of divisors of a positive integer $x$.
Define $$s(x):=\sigma(x)-x$$ to be the sum of the aliquot divisors of $x$, and define $$D(x):=2x-\sigma(x)$$ to be the deficiency of $x$.
Now, let $p$ be a prime number, and let $k$ be a positive integer.
Here is my question:
Is it true that $\gcd(s(p^k), D(p^k)) = 1$?
MY ATTEMPT
If $k=1$, then $s(p^k)=s(p)=\sigma(p)-p=(p+1)-p=1$, so that clearly $$\gcd(s(p), D(p))=1.$$
Now let $k>1$. We obtain $$D(p^k) = 2p^k - \sigma(p^k) = p^k - \bigg(\sigma(p^k) - p^k\bigg) = p^k - s(p^k)$$ so that we get $$\gcd(s(p^k), D(p^k)) = \gcd(s(p^k), p^k - s(p^k)) = \gcd(s(p^k), p^k)$$ $$= \gcd(\sigma(p^k) - p^k, p^k) = \gcd(\sigma(p^k), p^k) = 1.$$
Note that a proof for the assertion that $\gcd(\sigma(p^k),p^k)=1$ is standard material in undergraduate textbooks on (elementary) number theory. So essentially, I will be assuming this assertion without proof.
Is this proof correct?
As already mentioned in the comment, the only prime possibly both dividing $s(p^k)$ and $D(p^k)$ is $p$.
Hence it remains to show that $s(p^k)$ is not divisible by $p$.
$s(p^k)$ is the sum of the divisors of $p^k$ except $p^k$ itself. All the divisors except $1$ are divisible by $p$, hence the sum cannot be divisible by $p$ (The sum consists of $1$ and in the case $k>1$ of other summands divisible by $p$ , in the case $k=1$ , the sum is just $1$). This completes the proof.