, 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 $u_n \equiv -1 \pmod{3}$ with every $n \ge 2$. Similarly, $u_n$ is odd for all $n$. Also, $675 \equiv 1 \pmod{337}$ so $u_0 \equiv 1 \pmod{337}$ entails $u_1 \equiv -1 \pmod{337}$ and so on we also inductively $u_n \equiv -1 \pmod{337}, \forall n \ge 1$. From this it follows that $\text{gcd}(u_n, 2022) = 1, \forall n \ge 1$ and so
$$\varphi(2022u_n) = \varphi(2)\varphi(3)\varphi(337)\varphi(u_n) = 672\varphi(u_n).$$
Next, $675 \equiv 3 \pmod 7$ so $u_1 \equiv 3^2 \equiv 2 \pmod 7$ and inductively $u_n \equiv 2 \pmod 7, \forall n \ge 1$. Finally, notice that if $d = \text{gcd}(m, n)$ then
$$\varphi(mn) = \varphi(m)\varphi(n) \frac{d}{\varphi(d)} \ge \varphi(m)\varphi(n).$$
Hence one can conclude that
$$\varphi(2023u_n) \ge \varphi(7)\varphi(17^2)\varphi(u_n) = 1632\varphi(u_n) > 2 \cdot 672\varphi(u_n) = \varphi(2022u_n)$$
with every $n \in \mathbb{Z}^+$.