BxMO/EGMO Team Selection Test

The only value that satisfies is $n=1175$.

First, we note that we can rewrite the recursion as $$a_{n+1}=\frac{a_n^2-1+1}{a_n-1}=\frac{a_n^2-1}{a_n-1}+\frac{1}{a_n-1}=a_n+1+\frac{1}{a_n-1}.(1)$$ Since the difference of $a_{n+1}-a_n>1$, there is at most one $n$ that satisfies. Now we are going to show that $n=1175$ does indeed satisfy. First, the above gives that $$\begin{array}{rl}{a}_{1175}& =a_1+1174+\frac{1}{a_1-1}+\frac{1}{a_2-1}+\cdots +\frac{1}{a_{1174}-1}\\ & >850+1174\\ & =2024.\end{array}$$ More generally, it follows from (1) that $a_n\ge 849+n$. From this we deduce that $$\begin{array}{rl}\frac{1}{a_1-1}+\cdots +\frac{1}{a_{1174}-1}& \le \frac{1}{849}+\frac{1}{850}+\cdots +\frac{1}{2022}\\ & <51\cdot \frac{1}{849}+100\cdot \frac{1}{900}+200\cdot \frac{1}{1000}+400\cdot \frac{1}{1200}+400\cdot \frac{1}{1600}+23\cdot \frac{1}{2000}\\ & =\frac{17}{283}+\frac{1}{9}+\frac{1}{5}+\frac{1}{3}+\frac{1}{4}+\frac{23}{2000}\\ & =\frac{17}{283}+\frac{20+36+60+45}{180}+\frac{23}{2000}\\ & <\frac{11}{180}+\frac{161}{180}+\frac{3}{180}\\ & <1.\end{array}$$ Using this, we conclude that indeed $$ a_{1175} \le 850 + 1174 + \frac{1}{849} + \frac{1}{850} + \cdots + \frac{1}{2022} < 2024 + 1. \quad \square