Assume that there are finitely many odd prime number $p$ Such that there exist $n\in\mathbb{N}$ which $p|n^{2022}+2022$ Let it be $\{a_1,a_2,..,a_k\}\cup \{3,337\}$ where $a_1<a_2<…<a_k$ Since $((3)(337)a_1a_2…a_k,(a_1a_2…a_k)^{2022}+2022)=1$ there must exist odd prime $k\notin \{a_1a_2,..,a_k\} \cup \{3,337\} $ Where $k| (a_1a_2…a_k)^{2022}+2022$ a contradiction Hence, there exist infinitely many odd prime number $p$ Such that there exist $n\in\mathbb{N}$ which $p|n^{2022}+2022~~~(*)$ Let $p$ be an odd prime number $p$ such that there exist $n\in\mathbb{N}$ which $p|n^{2022}+2022$ and $p\notin\{3,337\}$ Let $q\in\mathbb{N}$ Assume that there exist $n\in\mathbb{N}$ which $p^q|n^{2022}+2022,~~$ Thus $n^{2022}\equiv -2022+rp^q(mod~p^{q+1})$ where $r\in\mathbb{N}$ We can clearly see that $(n,p)=1$ thus there exist $m\in\mathbb{N}$ Such that ${2022 \choose 1}n^{2021}(mp^q)\equiv -rp^q(mod~p^{q+1})$ Thus $(n+mp^q)^{2022}+2022\equiv n^{2022}+ {2022 \choose 1}n^{2021}(mp^q)+2022(mod~p^{q+1})$ $\Rightarrow (n+mp^q)^{2022}+2022\equiv n^{2022} -rp^q+2022 (mod~p^{q+1})$ $\Rightarrow p^{q+1}| (n+mp^q)^{2022}+2022$ By induction we’ll get that for all $q\in\mathbb{N}$ there exist $n\in\mathbb{N}$ Such that $p^q|n^{2022}+2022~~~(**)$ From $(*),(**)$ we can conclude that there exist infinitely many prime $p$ which there is a positive integer $n$ such that $p^{2022}$ divides $n^{2022}+2022~~~\square$ lmk if there is any mistakes