@below, was wondering whether that was the case

@parmenides51 don't you mean any arbitrary integer instead of an arbitrary integer?

parmenides51 wrote: It is known that for some integers $a_{2021},a_{2020},...,a_1,a_0$ the expression $$a_{2021}n^{2021}+a_{2020}n^{2020}+...+a_1n+a_0$$is divisible by $2021$ for an arbitrary integer $n$. Is it required that each of the numbers $a_{2021},a_{2020},...,a_1,a_0$ also divisible by $2021$? This is off topic, but why do you have greek on your pfp?

anthonyshinex wrote: @parmenides51 don't you mean any arbitrary integer instead of an arbitrary integer? It means any arbitrary integer n @above, cause I am from Greece

parmenides51 wrote: anthonyshinex wrote: @parmenides51 don't you mean any arbitrary integer instead of an arbitrary integer? It means any arbitrary integer n @above, cause I am from Greece ohh nice! Our school teaches us greek.

I don't think the solution above is correct. I believe the answer is no - let me know if you spot an error below. Set $2021=43\cdot 47$, and recall that $n^{42k+\ell}\equiv n^\ell\pmod{43}$ and $n^{46k'+\ell}\equiv n^\ell \pmod{47}$ for any $k,k',\ell,n$ by Fermat. With this, take first modulo $43$, and express $P$ as a polynomial $\alpha_{41}n^{41}+\cdots+\alpha_1 n+\alpha_0$ where \[ \alpha_i = \sum_{0\le j\le 2021, j\equiv i\pmod{42}} a_j,\quad\quad\text{for}\quad 0\le i\le 41. \]It is clear that one can choose $a_i$ such that while not all $a_i$ are zero modulo $43$, the resulting $\alpha_i$'s are; making $P$ divisible by $43$. Now do the same for $47$.

Choose $a_i$ to be divisible by $2021$ but $a_1$ and $a_{\phi(2021+1}$. Let $a_1 = -a_{\phi(2021) + 1}$ and since $n^{\phi{(2021)}+1} \equiv n \pmod{2021}, (n;2021)=1$ we get that $P(n)$ is divisible by $2021$ for all $n, (n;2021)=1$. WLOG $43|n$ and $47\nmid n$, than $n^{\phi{(2021)}+ 1} \equiv n^{\phi{(43)}+1} \equiv n \pmod{43}$, since $\phi(2021) = \phi{(43)}\phi{(47)}$.

Take the polynomial $$f(x)=\prod_{k=1}^{2021}(x+k)$$as a quick counterexample to necessity.