For what primes $p$ and positive integers $k$ is this algebraic expression divisible by $3$? My initial question is as is in the title:

For what prime $p$ and positive integers $k$ is this algebraic expression divisible by $3$?
$$A(p,k):=\dfrac{p^{2k+2} - 4p^{2k+1} + 6p^{2k} + 2p^{k+1} - 8p^k + 3}{2(p - 1)^2}$$

I would like to qualify that I am specifically interested in those values for $p$ and $k$ satisfying the congruence $p \equiv k \equiv 1 \pmod 4$.
MY ATTEMPT
Here, I will evaluate my expression for
$$p_1 = 5, k_1 = 1,$$
which gives $A(p_1,k_1)=9$ (which is divisible by $3$),
$$p_2 = 13, k_2 = 1,$$
which gives $A(p_2,k_2)=73$ (which is not divisible by $3$), and
$$p_3 = 17, k_3 = 1,$$
which gives $A(p_3,k_3)=129$ (which is divisible by $3$).
Here is my final question:

Will it be possible to consider the cases $p \equiv 1 \pmod 3$ and $p \equiv 2 \pmod 3$ separately, then use the Chinese Remainder Theorem afterwards?  (I know the concept, but I would have forgotten how to do that.)

CONJECTURE: If $p \equiv 2 \pmod 3$, then $3 \mid A(p,k)$.
Alas, this is where I get stuck.

Consider the point that the sum of coefficiant of numerator is 0, so for $p\equiv 1\bmod3$ the numerator is divisible by 3 for any k. For the case $p\equiv 2\bmod 3$ we have:
$p\equiv 2 \bmod 3\Rightarrow p^2\equiv 1 \bmod 3\Rightarrow p^{2(k+1)}\equiv 1 \bmod 3$
$p^{2k+1}=p^{2k}\cdot p\equiv(1\bmod 3)(2\bmod 3)\equiv 2\bmod 3\Rightarrow 4p^{2k+1}\equiv 8 \bmod 3\equiv 2 \bmod 3$
$6p^{2k}\equiv 0 \bmod 3$
$p^{k+1}p^k\cdot p\equiv (2^k \bmod 3)(2\bmod 3)\Rightarrow 2p^{k+1}\equiv 2^{k+2} \bmod 3$
$8p^k\equiv 2^{k+3} \bmod 3$
Summing these we get for numerator:
$r=1-2+0+2^{k+2}-2^{k+3}-3\equiv 2^{k+2}(1-2)-1\equiv -1-2^{k+2}\bmod 3$
$2^{k+1}=(3-1)^{k+1}$
So if $k=2m+1$ we have:
$r=-1 -3t+1=-3t\equiv 0 \bmod 3$
