As stated in the title. This is part of a homework assignment.
2026-04-30 05:05:17.1777525517
Show that there is no polynomial $f \in (\mathbb{Z}/100\mathbb{Z})$ satisfying f(1)=1 and f(11)=17
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If there is such an $f$, let $g(x)=f(x+1) -x-1$ (this will still be some polynomial). The given information about $f$ translates into $g(0)= 0$ and $g(10) = 6$. Can you see why no such $g$ can exist?