I came across this question in an Olympiad mathematics book:
If $f(x) \in \mathbb{Z}\left[x\right]$ and $f(\alpha)\equiv k\pmod{n}$ then prove $f(\alpha +mn)\equiv k \pmod{n}$ for every integer $m$.
This was my attempt at solving it:
Let take an expression of the form $$a_p(\alpha+mn)^p$$ Then this expression can be written as $a_p\alpha^p + p(mn)$ where $p(x) \in \mathbb{Z}\left[x\right]$. Now we can write $$f(x) = a_0 + a_1\alpha +\ldots+ a_n\alpha^n \Rightarrow f(\alpha+mn) = a_0 + a_1(\alpha+mn) +\ldots+ a_n(\alpha+mn)^n$$
Which can, in turn, be written as $$f(\alpha)+p(mn)$$
Now we know that $n \mid f(\alpha)-k$ and it is obvious that $n \mid p(mn)$. Hence we can say that $n \mid f(\alpha)+p(mn)-k$ which implies $f(\alpha +mn)\equiv k \pmod{n}$.
Is the proof correct? I showed it to my instructor and he was not convinced with the expansion of $a_p(\alpha+mn)^p$.