XIX Silk Road Mathematical Competition

By induction on $n$ it is easy to show that $a_n\ge n$ for any $n$. Suppose that there exists a positive integer $k$ such that $a_k>k$. Let's choose such positive integer $m$ that $m:2021!$ and $m>k$. Then for any $i=2,3,\dots ,2021$, $\text{GCD}(m,m+i)>1$. It follows from the problem statement that $\text{GCD}(m,a_{m+1})=1$. Since $\{a_n\}$ is strictly increasing and $a_k>k$, then $a_{m+1}>m+1$. Therefore, $m+2\le a_{m+1}\le m+2021$, but then $\text{GCD}(m,a_{m+1})>1$ — a contradiction.

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Alternative solution.

Let $b_n=a_n-n$ for each $n$. By induction on $n$ it is easy to show that $b_n\ge 0$ for any $n$. If $b_k>b_{k+1}$ for some $k$, then $$a_k-k>a_{k+1}-k-1⟹a_k+1>a_{k+1}⟹a_k\ge a_{k+1}$$ — a contradiction. Thus, the sequence $\{b_n\}$ is non-decreasing. On the other hand, it has an upper bound: $b_n=a_n-n\le 2020$. Hence, there exists such non-negative integer $k$ and a positive integer $t$ that $b_n=k$ for each $n\ge t$. So, for any $n\ge t$ $$\begin{array}{rl}{a}_{n+1}\mid n^3{a}_n-1& ⟹n+k+1\mid n^3(n+k)-1\\ & ⟹n+k+1\mid n^3(n+k)-1-(n+k+1)(n^3-n^2+n(k+1)-(k+1{)}^2)=(k+1{)}^3-1\\ & ⟹n+k+1\mid (k+1{)}^3-1.\end{array}$$ But this is only possible when $k=0$. Therefore, $b_n=0$ for each sufficiently large $n$, and thus for all $n$, i.e. $a_n=n$ for all $n$.