# 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.

• What's the source of this problem, please? – Gerry Myerson Feb 24 at 22:57
• How did you make that only observation? – Berci Feb 25 at 1:36
• @Gerry Myerson contest from my country – MathEnthusiast Feb 25 at 5:32
• Definitely not. It was featured on the maths olympiad on Saturday, I would never post ongoing problems – MathEnthusiast Feb 25 at 11:44
• I added a screenshot of the official paper @Jyrki Lahtonen – MathEnthusiast Mar 11 at 18:04

The problem statement in general is false if $$A$$ and $$B$$ do not commute. E.g. consider $$A=\pmatrix{1&-1\\ 2&-1}, \ B=\pmatrix{0&-1\\ 1&0}.$$ Then $$A^2=B^2=-I$$, so that $$\det(A)=\det(B)=1\ \text{ and } \ \frac14\det\left(A^2+B^2\right)=\frac14\det(-2I)=1.$$ Yet, as $$A^{2019}=(A^2)^{1009}A=(-I)^{1009}A=-A$$ and the analogous holds for $$B$$, we have \begin{aligned} \det\left(A^{2019}+B^{2019}\right) =\det(-A-B)=\det(A+B) =\det\pmatrix{1&-2\\ 3&-1} =5, \end{aligned} which is not divisible by $$4$$. One can easily verify that $$AB\ne BA$$ in this example.
Nonetheless, the problem statement is true if $$AB=BA$$. This can be proved easily be considering the eigenvalues of $$A,\ B$$ and $$A^2+B^2$$.
The other answer demonstrates that the result is false unless $$AB=BA$$; here's a proof in that case. Note that $$A^2+B^2=\dfrac12\left((A+B)^2+(A-B)^2\right)$$. Therefore, $$4=\det(A^2+B^2)=\frac14\det\left((A+B)^2+(A-B)^2\right),$$ i.e. $$\det\left((A+B)^2+(A-B)^2\right)=16.$$ But because $$\det(X+Y)=2\det X+2\det Y-\det(X-Y)$$ for any two $$2\times 2$$ matrices $$X,Y$$, we have $$\det\left((A+B)^2+(A-B)^2\right)=2\det(A+B)^2+2\det(A-B)^2-\det(4AB),$$ so by the condition, $$16 = \det(A+B)^2+\det(A-B)^2.$$ Now, $$A+B$$ and $$A-B$$ are both integer matrices, so their determinants squared are nonnegative perfect squares. The only two nonnegative perfect squares that sum to $$16$$ are $$0,16$$, therefore one of $$\det(A+B),\det(A-B)$$ is $$0$$ and the other is $$\pm 4$$. We are done because of the factorisations $$A^{2019}+B^{2019}=(A+B)(A^{2018}-A^{2017}B+\dots+B^{2018}),$$ $$A^{2019}-B^{2019}=(A-B)(A^{2018}+A^{2017}B+\dots+B^{2018}).$$
• Why does $(A+B)^2-(A-B)^2=4AB$? It is $2(AB+BA)$. And the final identities don't hold generally, unless you assume $AB=BA$. – egreg Mar 10 at 16:49
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 $$4$$.