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## IMC2019: Day 1, Problem 5
(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
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 \(\displaystyle 4\).
\(\displaystyle A^4 \equiv A^4+4A^2B^2+16B^4 = 2019I \mod{4} \) 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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