WakeUp wrote: Let $n$ be a positive integer and $f(x)=a_mx^m+\ldots + a_1X+a_0$, with $m\ge 2$, a polynomial with integer coefficients such that: a) $a_2,a_3\ldots a_m$ are divisible by all prime factors of $n$, b) $a_1$ and $n$ are relatively prime. Prove that for any positive integer $k$, there exists a positive integer $c$, such that $f(c)$ is divisible by $n^k$. Hensel Lemma !

my solution: I will use the following well known lemma: Hensel's lemma: $P(x)$ is a polynomial with integer coefficients and $p$ is a prime number if for an integer $x$:$p\mid P(x),p\nmid P(x)$ then for every natural number $k$ there is an integer $x$ such that $p^k\mid P(x)$. Let $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ note that since $(a_1,p_i)=1$ there is an integer $x$ such that $p_i\mid a_1x+a_0$ or $p_i\mid P(x)$ note that because $p_i\nmid f'(x)\equiv a_1\pmod{p_i}$ and there is an $x$ such that $p_i\mid f(x)$ by hensel's lemma we deduce that for there is an $x_i$ such that $p_i^{ka_i}\mid f(x_i)$ now using Chinese remainder theorem there is an $x$ such that $x\equiv x_i\pmod{p_i}$ so $n^k\mid f(x)$ DONE