Saudi Arabian IMO Booklet

Solution. Denote $b_n=\frac{a_n}{\text{rad}(a_n)}$. Since rad($a_n$) divides rad($a_{n+1}$) we have $b_{n+1}|b_n+1$. If there are indices $i<j$ with $b_i<2022<b_{i+1}$, we will be done by “continuity”. If, to the contrary, this does not happen, there are two possible cases. $b_n<2022$ for all $n$ big enough. Since $a_n$ increases indefinitely, then so does rad($a_n$), so at some moment rad($a_n$) receives a new prime $p>2022$. This means that $p\nmid a_n$ and $p|a_{n+1}=\frac{b_{n+1}}{b_n}{a}_n$, so $p|b_n+1$ and hence $b_n\ge 2022$, a contradiction. $b_n>2022$ for all $n$. We can assume WLOG that $b_0$ is the smallest term of the sequence $(b_n)$. Suppose that $b_{i+1}=b_i+1$ for all $0\le i<n$. Then $$\text{rad}(a_0)=\cdots =\text{rad}(a_{n-1})=R.$$ But for every prime $p\le n$ there is a multiple of $p$ among us sus $b_0,\dots ,b_{n-1}$, so $p|a_k$ for some $k$ and consequently $p|R$. Since not every prime divides $R$, there must be an index $n$ such that $b_n<b_{n-1}+1$, i.e. $b_{n-1}+1=d{b}_n$ for some $d>1$ and $\text{rad}(a_{n+1})=dR$, so $\text{gcd}(d,R)=1$. Recall that $b_0\le b_n=\frac{b_0+n}{d}$, which reduces to $n\ge (d-1)b_0$. By above, this means that all primes up to $(d-1)b_0$ divide $R$, but $d$ does not divide $R$, so $d>(d-1)b_0$, which is impossible. $\square$