Problem: Q ( z ) Q ( z ) R ( z ) R ( z )
z 2021 + 1 = ( z 2 + z + 1 ) Q ( z ) + R ( z ) z 2 0 2 1 + 1 = ( z 2 + z + 1 ) Q ( z ) + R ( z )
and the degree of R R R ( z ) R ( z )
Answer Choices:
A. β z β z β 1 β 1 2021 2 0 2 1 z + 1 z + 1 2 z + 1 2 z + 1 Solution:
Answer (A): Note that
z n + 3 + 1 β ( z n + 1 ) = z n ( z 3 β 1 ) = z n ( z β 1 ) ( z 2 + z + 1 ) . z n + 3 + 1 β ( z n + 1 ) = z n ( z 3 β 1 ) = z n ( z β 1 ) ( z 2 + z + 1 ) .
Therefore z n + 3 + 1 z n + 3 + 1 z n + 1 z n + 1 z 2 + z + 1 z 2 + z + 1 2021 β‘ 2 ( β β 3 ) 2 0 2 1 β‘ 2 ( m o d 3 ) z 2021 + 1 z 2 0 2 1 + 1 z 2 + z + 1 z 2 + z + 1 z 2 + 1 z 2 + 1 z 2 + z + 1 z 2 + z + 1 β z β z
OR
The solutions of z 2 + z + 1 = 0 z 2 + z + 1 = 0 z = β 1 Β± i 3 2 z = 2 β 1 Β± i 3 β β
R ( β 1 + i 3 2 ) = R ( cis β‘ 2 Ο 3 ) = ( cis β‘ 2 Ο 3 ) 2021 + 1 = ( cis β‘ 4042 Ο 3 ) + 1 = ( cis β‘ 4 Ο 3 ) + 1 = 1 β i 3 2 R ( 2 β 1 + i 3 β β ) β = R ( c i s 3 2 Ο β ) = ( c i s 3 2 Ο β ) 2 0 2 1 + 1 = ( c i s 3 4 0 4 2 Ο β ) + 1 = ( c i s 3 4 Ο β ) + 1 = 2 1 β i 3 β β β
and
R ( β 1 β i 3 2 ) = R ( cis β‘ 4 Ο 3 ) = ( cis β‘ 4 Ο 3 ) 2021 + 1 = ( cis β‘ 8084 Ο 3 ) + 1 = ( cis β‘ 2 Ο 3 ) + 1 = 1 + i 3 2 . R ( 2 β 1 β i 3 β β ) β = R ( c i s 3 4 Ο β ) = ( c i s 3 4 Ο β ) 2 0 2 1 + 1 = ( c i s 3 8 0 8 4 Ο β ) + 1 = ( c i s 3 2 Ο β ) + 1 = 2 1 + i 3 β β . β
Because R ( z ) R ( z ) R ( z ) = β z R ( z ) = β z z 2 + z + 1 = 0 z 2 + z + 1 = 0 R ( z ) = β z R ( z ) = β z
The problems on this page are the property of the MAA's American Mathematics Competitions