If the polynomial `z^2 + z + 1 ` would be linear, it would be simple to find the remainder using the remainder theorem. We can factor it under the complex numbers: `z^2 + z + 1 = ( z - z_1 ) ( z - z_2 ) ,` where

`z_( 1 , 2 ) = ( -1 +- i sqrt ( 3 ) ) / 2 = e^( +-( 2 pi i ) / 3 ) .`

Substitute `z_1 ` and `z_2 ` into the equation, and recall that `R ( x ) = ax + b :`

`z_1^2021 + 1 = a z_1 + b , ` `z_2^2021 + 1 = a z_2 + b .`

It is a linear system for `a ` and `b ,` the solutions are

`a = ( z_2^2021 - z_1^2021 ) / (z_2 - z_1) ,` `b = ( z_2 z_1^2021 - z_1 z_2^2021 ) / (z_2 - z_1) + 1 .`

To make these expressions look better, compute the powers:

`z_1^2021 = e^( ( 2 pi i ) ( 2021 / 3 ) ) = e^( ( 2 pi i ) ( 674 - 1 / 3 ) ) = e^( - ( 2 pi i ) / 3 ) = z_2 , ` so `z_2^2021 = z_1 .`

Now we see `a = -1 ` and `b = z_2 + z_1 + 1 = 0 , ` so `R ( z ) = -z .`

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