If $p^k m$ is deficient (with $p$ a prime number satisfying $\gcd(p,m)=1$), then what is the sign/residue of $D(p^k)D(m) - 2s(p^k)s(m)$ (modulo $3$)? Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.  Denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the aliquot sum of $x$ by $s(x)=\sigma(x)-x$.  Finally, denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Let $p \equiv 1 \pmod 4$ be a prime number satisfying $\gcd(p,m)=1$, where $m$ is a positive integer.  Let $k \equiv 1 \pmod 4$ be a positive integer.  Suppose that $p^k m$ is deficient.  Then we know (by definition) that $I(p^k m) < 2$.
My initial question is as is in the first part of the title:

If $p^k m$ is deficient (with $p$ a prime number satisfying $\gcd(p,m)=1$), then what is the sign of $D(p^k)D(m) - 2s(p^k)s(m)$?

Note that
$$p^k m \text{ is deficient } \iff I(p^k m) < 2 \iff 2I(p^k m) + 2 < I(p^k m) + 4$$
$$\iff 4 - 2(I(p^k) + I(m)) + I(p^k m) > 2I(p^k m) - 2(I(p^k) + I(m)) + 2.$$
But we obtain
$$4 - 2(I(p^k) + I(m)) + I(p^k m) = 4 - 2(I(p^k) + I(m)) + I(p^k)I(m) \text{ since } \gcd(p,m)=1$$
and
$$2I(p^k m) - 2(I(p^k) + I(m)) + 2 = 2I(p^k)I(m) - 2(I(p^k) + I(m)) + 2 \text{ since } \gcd(p,m)=1,$$
which can be rewritten as
$$4 - 2(I(p^k) + I(m)) + I(p^k)I(m) = (2 - I(p^k))(2 - I(m))$$
and
$$2I(p^k)I(m) - 2(I(p^k) + I(m)) + 2 = 2(I(p^k) - 1)(I(m) - 1).$$
We therefore have the inequality
$$(2 - I(p^k))(2 - I(m)) > 2(I(p^k) - 1)(I(m) - 1).$$
Multiplying both sides by $p^k m$, we obtain
$$(2p^k - \sigma(p^k))(2m - \sigma(m)) = D(p^k)D(m) > 2s(p^k)s(m) = 2(\sigma(p^k) - p^k)(\sigma(m) - m),$$
which seems to show that
$$\Delta = D(p^k)D(m) - 2s(p^k)s(m) > 0.$$
ANOTHER WAY TO DETERMINE THE SIGN OF $\Delta$
As before, let
$$\Delta = D(p^k)D(m) - 2s(p^k)s(m).$$
Since $p$ is prime, we obtain
$$D(p^k) = \dfrac{p^{k+1} - 2p^k + 1}{p - 1}$$
and
$$s(p^k) = \dfrac{p^k - 1}{p - 1}.$$
We obtain the expression
$$\Delta = D(p^k)D(m) - 2s(p^k)s(m)$$
$$= \bigg(\dfrac{p^{k+1} - 2p^k + 1}{p - 1}\bigg)\cdot(2m - \sigma(m)) + \dfrac{2 - 2p^k}{p - 1}\cdot(\sigma(m) - m).$$
We consider the coefficients of $\sigma(m)$ and $m$ separately.
The coefficient of $\sigma(m)$ is:
$$-\bigg(\dfrac{p^{k+1} - 2p^k + 1}{p - 1}\bigg) + \dfrac{2 - 2p^k}{p - 1} = \dfrac{-p^{k+1} + 1}{p - 1} = -\sigma(p^k),$$
while the coefficient of $m$ is:
$$2\bigg(\dfrac{p^{k+1} - 2p^k + 1}{p - 1}\bigg) - \dfrac{2 - 2p^k}{p - 1} = \dfrac{2p^{k+1} - 4p^k + 2 - 2 + 2p^k}{p - 1} = \dfrac{2p^{k+1} - 2p^k}{p - 1} = 2p^k.$$
Hence,
$$\Delta = D(p^k)D(m) - 2s(p^k)s(m) = -\sigma(p^k)\sigma(m) + 2p^k m > 0,$$
since $I(p^k m) < 2$ implies that $I(p^k)I(m) < 2$, giving
$$\sigma(p^k)\sigma(m) < 2p^k m.$$
Here is my final question:

Can the residue of $\Delta$ modulo $3$ be determined?

MY ATTEMPT TO ANSWER THE FINAL QUESTION
Obviously, since $p \equiv 1 \pmod 4$ and $p$ is prime, then $p \neq 3$.  Hence, either $p \equiv 1 \pmod 3$ or $p \equiv 2 \pmod 3$.  But $(p+1) \mid \sigma(p^k)$ (since $p \equiv k \equiv 1 \pmod 4$), whence we obtain $\sigma(p^k) \equiv 0 \pmod 3$ if $p \equiv 2 \pmod 3$.  (I am not sure, whether there are other instances, for example when $p \equiv 1 \pmod 3$ and $k$ is restricted to some congruence class modulo $3$, that will make $\sigma(p^k) \equiv 0 \pmod 3$ too.)
Note that the considerations for the previous paragraph are only for the term
$-\sigma(p^k)\sigma(m)$.  I have not even began to consider the contribution of the term $2p^k m$.
Alas, this is where I get stuck!
 A: This answer proves the following claims :
Claim 1 : If $m\equiv 2\pmod 3$, then $\sigma(m)\equiv 0\pmod 3$.
Claim 2 : $$\Delta\stackrel{\text{mod}\ 3}\equiv\begin{cases} 2m-(k+1)\sigma(m)&\text{if $\ p\equiv 1\pmod 3$}\\\\ m&\text{if $\ p\equiv 2\pmod 3$}\end{cases}$$

Claim 1 : If $m\equiv 2\pmod 3$, then $\sigma(m)\equiv 0\pmod 3$.
Proof :
Let $q$ be a prime number and $s$ be a positive integer. Then, one has
$$\sigma(q^s)=1+q+\cdots +q^s\stackrel{\text{mod}\ 3}\equiv\begin{cases}1&\text{if $\ q\equiv 0\pmod 3$}\\\\1&\text{if $\ q\equiv 1,s\equiv 0\pmod 3$}\\\\2&\text{if $\ q\equiv 1,s\equiv 1\pmod 3$}\\\\0&\text{if $\ q\equiv 1,s\equiv 2\pmod 3$}\\\\1&\text{if $\ q\equiv 2\pmod 3$, $s$ even}\\\\0&\text{if $\ q\equiv 2\pmod 3$, $s$ odd}\end{cases}$$
So, if $m\equiv 2\pmod 3$, then there is a pair of positive integers $(q,s)$ such that $q$ is a prime number with $q^s||m,q\equiv 2\pmod 3$ and $s$ odd. So, one has $\sigma(m)=\sigma(q^s)\sigma\bigg(\dfrac{m}{q^s}\bigg)\equiv 0\pmod 3$.$\quad\blacksquare$

Claim 2 : $$\Delta\stackrel{\text{mod}\ 3}\equiv\begin{cases} 2m-(k+1)\sigma(m)&\text{if $\ p\equiv 1\pmod 3$}\\\\ m&\text{if $\ p\equiv 2\pmod 3$}\end{cases}$$
Proof :
If $p\equiv 1\pmod 3$, then since $\sigma(p^k)\equiv k+1\pmod 3$, one has $$\Delta = 2p^k m-\sigma(p^k)\sigma(m)\equiv 2m-(k+1)\sigma(m)\pmod 3.$$
If $p\equiv 2\pmod 3$, then since $\sigma(p^k)\equiv 0\pmod 3$, one has $$\Delta  = 2p^k m-\sigma(p^k)\sigma(m)\equiv -(-1)^{k} m\equiv m\pmod 3.\quad\blacksquare$$
