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

Representing the function ${\mathbb{Z}}_{9}\to {\mathbb{Z}}_{9}$, $f\left(0\right)=1$, $f\left(1\right)=\dots =f\left(8\right)=0$ as a polynomial in ${\mathbb{Z}}_{9}\left[x\right]$
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Dobricap
There is no such polynomial q.
We show this by contradiction. Assume that $q={a}_{k}{x}^{k}+\dots +{a}_{0}{x}^{0}\in {\mathbb{Z}}_{9}\left[x\right]$ is a polynomial with $q\left(0\right)=1$ and $q\left(x\right)=0$ otherwise.
Since ${0}^{2}={3}^{2}={6}^{2}=0$ in ${\mathbb{Z}}_{9}$, we get
$-1=f\left(3\right)-f\left(0\right)=\left({a}_{k}{3}^{k}+{a}_{k-1}{3}^{k-1}+\dots +{a}_{1}3+{a}_{0}\right)-\left({a}_{k}{0}^{k}+{a}_{k-1}{0}^{k-1}+\dots +{a}_{1}0+{a}_{0}\right)=3{a}_{1}.$
But in ${\mathbb{Z}}_{9}$, -1 is not a multiple of 3, so this is a contradiction.