Representing the function ZZ_9 => ZZ_9, f(0)=1, f(1)=…=f(8)=0 as a polynomial in ZZ_9[x]

Josiah Owens 2022-10-21 Answered
Representing the function Z 9 Z 9 , f ( 0 ) = 1, f ( 1 ) = = f ( 8 ) = 0 as a polynomial in Z 9 [ x ]
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Dobricap
Answered 2022-10-22 Author has 14 answers
There is no such polynomial q.
We show this by contradiction. Assume that q = a k x k + + a 0 x 0 Z 9 [ x ] is a polynomial with q ( 0 ) = 1 and q ( x ) = 0 otherwise.
Since 0 2 = 3 2 = 6 2 = 0 in Z 9 , we get
1 = f ( 3 ) f ( 0 ) = ( a k 3 k + a k 1 3 k 1 + + a 1 3 + a 0 ) ( a k 0 k + a k 1 0 k 1 + + a 1 0 + a 0 ) = 3 a 1 .
But in Z 9 , -1 is not a multiple of 3, so this is a contradiction.
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