IMC 2019 · Day 1 · P5

Hint: Consider the determinants modulo 4.

Remark. The proposed solution was more complicated and involved; during the contest it turned out that a signficantly simplified solution exists — which we now provide below.

We show that there are no such matrices.

Notice that $A^4 + 4 A^2 B^2 + 16 B^4$ can factorized as $$A^4 + 4 A^2 B^2 + 16 B^4 = (A^2 + 2AB + 4B^2)(A^2 - 2AB + 4B^2).$$ Let $C = A^2 + 2AB + 4B^2$ and $D = A^2 - 2AB + 4B^2$ be the two factors above. Then $$\det C \cdot \det D = \det(CD) = \det(A^4 + 4 A^2 B^2 + 16 B^4) = \det(2019 I) = 2019^n.$$ The matrices $C, D$ have integer entries, so their determinants are integers. Moreover, from $C \equiv D \pmod 4$ we can see that $$\det C \equiv \det D \pmod 4.$$ This implies that $\det C \cdot \det D \equiv (\det C)^2 \pmod 4$, but this is a contradiction because $2019^n \equiv 3 \pmod 4$ is a quadratic nonresidue modulo 4.

Notice that $$A^4 \equiv A^4 + 4 A^2 B^2 + 16 B^4 = 2019 I \mod 4$$ so $$(\det A)^4 = \det A^4 \equiv \det(2109 I) = 2019^n \pmod 4.$$ But $2019^n \equiv 3$ is a quadratic nonresidue modulo 4, contradiction.