Prove that $\det(A^{2019} +B^{2019} )$ and $\det(A^{2019} -B^{2019} )$ are divisible by $4$ 
Let $A, B \in M_2(\mathbb{Z})$ so that $$\det A=\det B=\frac{1}{4} \det(A^2+B^2)=1$$ Prove that $\det(A^{2019} +B^{2019} )$ and $\det(A^{2019} -B^{2019} )$ are divisible by $4$.     

The only observation I have made is that $\det(A^{2019} +B^{2019}) +\det(A^{2019} - B^{2019} )=4$ so suffice it to prove that one of them is divisible by $4$.
EDIT : This was featured on the regional stage of the maths olympiad in Romania on Saturday, so it is an actual problem, not something I came up with. It is relevant to me because I am preparing for the next stage.
EDIT : $AB=BA$ indeed, I forgot to include it when I posted the problem and I am sorry for this.
Since there are people in the comments eager to see the official paper, here it is :
https://imgur.com/a/z8v3VZp
As you can see, this contest took place on the 24th of February(I know we have a pretty strange date format, but this is what 24.02. 2019 means) and it wasn't an online competition, so I really was honest when I said I was not cheating. 

The OP has clarified that $AB=BA$. Then the problem can be solved easily by applying the identity
$$
\det(X+Y)+\det(X-Y)=2\left(\det(X)+\det(Y)\right)\tag{1}
$$
for $2\times2$ matrices twice. First, by putting $(X,Y)=(A^2,B^2)$ into $(1)$, we get
$$
4+\det\left[(A+B)(A-B)\right]=4.
$$
Therefore at least one of $A+B$ or $A-B$ is singular. Since they are factors of $A^{2019}+B^{2019}$ and $A^{2019}-B^{2019}$ respectively, one of $\det(A^{2019}+B^{2019})$ and $\det(A^{2019}-B^{2019})$ is $0$. Second, by putting $(X,Y)=(A^{2019},B^{2019})$ into $(1)$, we get
$$
\det(A^{2019}+B^{2019})+\det(A^{2019}-B^{2019})=4.
$$
Hence the other determinant is $4$.
