SAUDI ARABIAN IMO Booklet 2023

The answer is No. Suppose by contradiction that there is such a sequence. First, we will prove by induction that $x_n>2$ for every $n\ge 2$. One can check with $n=2$, then assume that the assertion is true for $n=k\ge 2$, i.e. $x_k>2$ then $$\frac{2x_k^2+2}{x_k+3}-2=\frac{2x_k^2-2x_k-4}{x_k+3}=\frac{2(x_k-2)(x_k+1)}{x_k+3}>0.$$ So $x_{k+1}>\frac{2x_k^2+2}{x_k+3}>2$. Therefore, the assertion is also true for $n=k+1$ and it is also true for all $n\ge 2$. Hence, we have $$x_{n+1}-x_n\ge \frac{2x_n^2+2}{x_n+3}-x_n=\frac{(x_n-1)(x_n-2)}{x_n+3}>0,$$ which implies that $(x_n)$ increases strictly. On the other hand, we have $$\frac{2x_n+2}{x_n+3}<2,\forall n>0⟹x_{n+1}<2+2023=2025,\forall n\ge 1$$ so $(x_n)$ is upper bounded by $2025$. Hence, $(x_n)$ has a finite limit, called by $L$ and $2<L\le 2025$. Substituting in the initial condition, we have $$\frac{2L^2+2}{L+3}\le L\iff 1\le L\le 2,$$ this contradiction shows that there is no sequence $(x_n)$ satisfying the condition. $\square$