SAUDI ARABIAN IMO Booklet 2023

Notice that $2023=7\cdot 17^2$ and $2022=2\cdot 3\cdot 337$. Put $u_n=a^{2n}+a^{-2n}$ then it is easy to see $u_{n+1}=u_n^2-2$ for every $n\ge 0$. We have $u_0=a+\frac{1}{a}=675$ divisible by $3$ so $u_1$ divided by $3$ leaves $1$, then $u_2$ divides $3$ with remainder $-1$ and inductively $u_n\equiv -1(\mathrm{mod}3)$ with every $n\ge 2$. Similarly, $u_n$ is odd for all $n$. Also, $675\equiv 1(\mathrm{mod}337)$ so $u_0\equiv 1(\mathrm{mod}337)$ entails $u_1\equiv -1(\mathrm{mod}337)$ and so on we also inductively $u_n\equiv -1(\mathrm{mod}337),\forall n\ge 1$. From this it follows that $\text{gcd}(u_n,2022)=1,\forall n\ge 1$ and so $$\phi (2022u_n)=\phi (2)\phi (3)\phi (337)\phi (u_n)=672\phi (u_n).$$ Next, $675\equiv 3(\mathrm{mod}7)$ so $u_1\equiv 3^2\equiv 2(\mathrm{mod}7)$ and inductively $u_n\equiv 2(\mathrm{mod}7),\forall n\ge 1$. Finally, notice that if $d=\text{gcd}(m,n)$ then $$\phi (mn)=\phi (m)\phi (n)\frac{d}{\phi (d)}\ge \phi (m)\phi (n).$$ Hence one can conclude that $$\phi (2023u_n)\ge \phi (7)\phi (17^2)\phi (u_n)=1632\phi (u_n)>2\cdot 672\phi (u_n)=\phi (2022u_n)$$ with every $n\in {\mathbb{Z}}^{+}$.