I tried using the method provided by Mark here but I can't seem to come up with a solution for $\mathbb{Z}_8$.
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Is the polynomial required to be monic? – Ben Grossmann Sep 16 '20 at 19:12
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@BenGrossmann No, that is not a requirement. – Jeremiah Sep 16 '20 at 19:16
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Going in the direction of the link you provided, what about $p(x)=4x(x+1)$?
mathcounterexamples.net
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Yes, this certainly seems to work. What was your motivation for constructing it in this way? – Jeremiah Sep 16 '20 at 19:18
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1Either $4x$ or $4(x+1)$ is equal to zero depending on whether $x$ is odd or even. – mathcounterexamples.net Sep 16 '20 at 19:19
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@Jeremiah I added an answer explaining why the linked method doesn't work and, further, how to discover the sought polynomial from this faiilure. – Bill Dubuque Sep 16 '20 at 20:18
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The linked method of iterating the Factor Theorem to get multiple factors from various roots works fine in fields (and domains), but generally it fails in non-domains like $\,\Bbb Z_8,\,$ where $\,4\cdot 2 = 0.\,$ Indeed, as explained here we can deduce that $\,f(a) = 0 = f(b)$ $\Rightarrow f(x) = (x-a)(x-b)g(x)\,$ only when $\,a-b\,$ is cancellable.
Let's see how it breaks. $\,f(0) = 0\,\Rightarrow\, f = x\, g.\,$ $\,0 = f(1) = 1\cdot g(1)\, $ so $\, g= (x\!-\!1)h,\,$ so $\,f = x(x\!-\!1)h.\,$ $\,0 = f(2) = 2(1)h(2),\,$ but we can't conclude $\,h(2)=0\,$ because $2$ is not cancellable in $\,\Bbb Z_8$. But we can infer $\,8\mid 2h(2)\,\Rightarrow\,4\mid h(2).\,$ Taking $\,h = 4\,$ does the job, i.e. $\,8\mid 4x(x\!-\!1)\,$ by $\,2\mid x(x\!-\!1),\,$ for all $\,x.$
Bill Dubuque
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