Prove that $p(x)$ has no integer zeros.

Question

Suppose $p(x)$ is a polynomial over $\mathbb Z$ such that there exists a positive integer $k$ for which none of the integers $p(1), p(2), \ldots , p(k)$ is divisible by $k$. Prove that $p(x)$ has no integer zeros.

My attempts:

I tried to prove it with contradiction. So I assumed that it has an integer zero $\alpha$. With it I got $\alpha \gt k$ or $\alpha \lt 0$, $a_0 \neq 0$, $k\nmid a_0$ and $\alpha \mid a_0$ here $a_0$ is the constant term of polynomial. But I was not able to get a contradiction. I would really appreciate any hints or help on this question. Thank you! for your help.

Answer 1

Suppose it is not true. Then there is $b\in \mathbb{Z}$ such that $p(b)=0$. Let $a:=b\pmod k< k$. Then we have $$k\mid a-b \mid p(a)-p(b) = p(a)$$

So if $a>0$ we have a contradiction. If $a=0$ then $k|b$ and since $b-k\mid p(b)-p(k)$ we have again $k\mid p(k)$ a contradiction again.

Answer 2

Thank you @Hw Chu for the hint. This answer is based on @Hw Chu's Hint.

$p(n)\equiv p(n + mk)\mod k$ for any $n$ and $m$ in $\mathbb Z$. So $$p(\alpha)\equiv p(\alpha + mk)\mod k \\ 0\equiv p(\alpha + mk)\mod k$$ $k\mid p(\alpha + mk)$ We know that $k\nmid \alpha \implies \alpha = kq + r$ where $0\lt r\lt k$ $$k\mid p(\alpha + mk)\\ k\mid p(kq + r + mk)\\ \text{taking} \;m=-q\\ k\mid p(kq + r - kq)\\ k\mid p(r)$$ which is a contradiction. Hence $p(x)$ has no integer zeros.