Exercise 2.2.3 (First examples, part 3)

Question

If a polynomial congruence $f(x) \equiv 0 \pmod m$ has $m$ solutions, prove that any integer whatsoever is a solution.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
The congruence $f(x) \equiv 0 \pmod m$ has $m$ solutions if and only if the equation $f(u) = 0$ has $m$ solutions $u \in \Z/m\Z$. Since $\Z/m\Z$ has $m$ elements, 
$$\forall u \in \Z/m\Z,\ f(u) = 0.$$
Therefore, for all $x \in \Z$, $f([x]_m) = 0$, thus $f(x) \equiv 0 \pmod m$.
\end{proof}

Exercise 2.2.3 (First examples, part 3)

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Exercise 2.2.3 (First examples, part 3)

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