International Mathematics Competition for University Students 2019

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IMC 2020
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IMC2019: Day 1, Problem 5

Problem 5. Determine whether there exist an odd positive integer $\displaystyle n$ and $\displaystyle n\times n$ matrices $\displaystyle A$ and $\displaystyle B$ with integer entries, that satisfy the following conditions:

(1) $\displaystyle \det(B)=1$;

(2) $\displaystyle AB=BA$;

(3) $\displaystyle A^4+4A^2B^2+16B^4=2019I$.

(Here $\displaystyle I$ denotes the $\displaystyle n\times n$ identity matrix.)

Proposed by Orif Ibrogimov, ETH Zurich and National University of Uzbekistan

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.

Solution 1. We show that there are no such matrices.

Notice that $\displaystyle A^4+4A^2B^2+16B^4$ can factorized as

$\displaystyle A^4+4A^2B^2+16B^4 = (A^2+2AB+4B^2)(A^2-2AB+4B^2).$

Let $\displaystyle C=A^2+2AB+4B^2$ and $\displaystyle D=A^2-2AB+4B^2$ be the two factors above. Then

$\displaystyle \det C \cdot \det D = \det (CD) = \det(A^4+4A^2B^2+16B^4) = \det(2019I) = 2019^n.$

The matrices $\displaystyle C,D$ have integer entries, so their determinants are integers. Moreover, from $\displaystyle C\equiv D\pmod 4$ we can see that

$\displaystyle \det C \equiv \det D \pmod{4}.$

This implies that $\displaystyle \det C\cdot \det D \equiv (\det C)^2 \pmod4$, but this is a contradiction because $\displaystyle 2019^n\equiv3\pmod{4}$ is a quadratic nonresidue modulo $\displaystyle 4$.

Solution 2. Notice that

$\displaystyle A^4 \equiv A^4+4A^2B^2+16B^4 = 2019I \mod{4}$

so

$\displaystyle (\det A)^4 = \det A^4 \equiv \det (2109I) = 2019^n \pmod{4}.$

But $\displaystyle 2019^n\equiv3$ is a quadratic nonresidue modulo $\displaystyle 4$, contradiction.

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