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 $s(x)=\sigma(x)-x$. (Note that, trivially, we have the identity $D(x)+s(x)=x$.) 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), and let $k \equiv 1 \pmod 4$ be a positive integer. Hereinafter, let $p^k m$ be a deficient number.
In a recent post, I proved that $$D(p^k)D(m) - 2s(p^k)s(m) = 2p^k m - \sigma(p^k)\sigma(m).$$
Now consider the identity $$p^k m = (D(p^k) + s(p^k))(D(m) + s(m)).$$ This can be rewritten as $$p^k m = D(p^k)D(m) + s(p^k)s(m) + D(p^k)s(m) + s(p^k)D(m),$$ or as $$p^k m = 3s(p^k)s(m) + \bigg(D(p^k)D(m) - 2s(p^k)s(m)\bigg) + D(p^k)s(m) + s(p^k)D(m).$$ Equivalently, we have $$p^k m = 3s(p^k)s(m) + \bigg(2p^k m - \sigma(p^k)\sigma(m)\bigg) + \bigg(\dfrac{p^{k+1} - 2p^k + 1}{p - 1}\bigg)\cdot(\sigma(m) - m) + \bigg(\dfrac{p^k - 1}{p - 1}\bigg)\cdot(2m - \sigma(m))$$ $$= 3s(p^k)s(m) + m\cdot\Bigg(2p^k - \bigg(\dfrac{p^{k+1} - 2p^k + 1}{p - 1}\bigg) + 2\cdot\bigg(\dfrac{p^k - 1}{p - 1}\bigg)\Bigg)$$ $$+ \sigma(m)\cdot\Bigg(\bigg(\dfrac{1 - p^{k+1}}{p - 1}\bigg) + \bigg(\dfrac{p^{k+1} - 2p^k + 1}{p - 1}\bigg) - \bigg(\dfrac{p^k - 1}{p - 1}\bigg)\Bigg),$$ whence we obtain $$p^k m = 3s(p^k)s(m) + m\cdot\bigg(\dfrac{p^{k+1} + 2p^k - 3}{p - 1}\bigg) + \sigma(m)\cdot\Bigg(-3\bigg(\dfrac{p^k - 1}{p - 1}\bigg)\Bigg)$$ $$= 3s(p^k)s(m) - 3s(p^k)\sigma(m) + m\cdot\bigg(\dfrac{p^{k+1} + 2p^k - 3}{p - 1}\bigg).$$
Here is my question:
For what primes $p \equiv 1 \pmod 4$ and positive integers $k \equiv 1 \pmod 4$ is the algebraic expression $$C(p,k):=\dfrac{p^{k+1} + 2p^k - 3}{p - 1}$$ divisible by $3$?
MY ATTEMPT
Like in the hyperlinked question, since $p$ is a prime satisfying $p \equiv 1 \pmod 4$, then $p \neq 3$. Therefore, either $p \equiv 1 \pmod 3$ or $p \equiv 2 \pmod 3$. (Note that $k$ is odd.)
Case 1: When $p \equiv 1 \pmod 3$, the numerator is divisible by $3$, while the denominator is also divisible by $3$. Hence, I cannot conclude whether $3 \mid C(p,k)$ or not, under this case.
Case 2: When $p \equiv 2 \pmod 3$, the numerator is not divisible by $3$. Hence, I can conclude that $3 \nmid C(p,k)$, under this case.
I suspect an approach similar to that done in this answer to a tangentially related question would do the trick.
Alas, this is where I get stuck, as I do not possess the mathematical prowess needed to conclusively settle Case 1.
Note that $k$ is odd and $$C(p,k)=\frac{(p^{k+1}-p^k)+3(p^k-1)}{p-1}$$ $$=p^k+3\cdot \frac{p^k-1}{p-1},$$ where the second term is divisible by $3$, hence the expression is never divisible by $3$, since $p\equiv 1~({\rm mod~}4).$