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1 solution
Let R ( x ) = ( x − 2 0 1 8 ) Q ( x ) . Then R is also a rational function and R ( x ) = R ( x + 2 0 1 8 ) . Nonconstant rational functions are not periodic, so R is constant. Since R ( 1 ) = − 2 0 1 7 , we get ( x − 2 0 1 8 ) Q ( x ) = − 2 0 1 7 , so Q ( x ) = 2 0 1 8 − x 2 0 1 7 . Then Q ( 2 0 1 7 ) = 2 0 1 7 .
(To see why nonconstant rational functions aren't periodic, consider R ( x ) = g ( x ) / h ( x ) , and suppose R is periodic. If g is nonconstant, then the number of zeroes of R ( x ) is finite and nonzero, which is impossible for a periodic function; ditto for the number of zeroes of the periodic function 1 / R ( x ) if h is nonconstant.)