### 2020 Silk Road #1

Given a strictly increasing infinite sequence of natural numbers $ a_1, $ $ a_2, $ $ a_3, $ $ \ldots $. It is known that $ a_n \leq n + 2020 $ and the number $ n ^ 3 a_n - 1 $ is divisible by $ a_ {n + 1} $ for all natural numbers $ n $. Prove that $ a_n = n $ for all natural numbers $ n $.

Notee that $a_{n}\ge a_{n-1}\ge{\dots}\ge{a_{1}+(n-1)}\ge{n}$. Now for the sake of contradiction, assume $\exists{i\in{\mathbb{N}}}$ for which $a_{i}>i$. The condition $a_{n+1}\mid{n^3a_{n}-1}$ implies ${\gcd(a_{n+1},n)=1}$, for all $n\in{\mathbb{N}}$. Notice that if we take $j$ such that $2021!\mid{j}$ and $j>i$ , then $\gcd(j,j+t)>1$ for all $2\le{t}\le{2021}$ . Since the sequence is increasing and $a_{i}>i$ and $j>i$ , we have $a_{j+1}>j+1$. Then, $j+2\le{a_{j+1}}\le{j+2021}$ and so $1=\gcd(a_{j+1},j)>1$ , which is a contradiction. Thus, $a_n=n$, as desired. $\square$

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