Among the statements :

(S1) : $2023^{2022}-1999^{2022}$ is divisible by 8

(S2) : $13(13)^{n}-12 n-13$ is divisible by 144 for infinitely many $n \in \mathbb{N}$

We have, $S_1$ : $(2023)^{2022}-(1999)^{2022}$ <br/><br/>$$ \begin{aligned} & =(1999+24)^{2022}-(1999)^{2022}={ }^{2022} C_0(1999)^{2022}(24)^0 \\\\ & +{ }^{2022} C_1(1999)^{2021}(24)^1+{ }^{2022} C_2(1999)^{2020}(24)^2 \\\\ & +\ldots-(1999)^{2022} \\\\ & ={ }^{2022} C_1(1999)^{2021}(24)+{ }^{2022} C_2(1999)^{2022}(24)^2 \\\\ & =24\left({ }^{2022} C_1(1999)^{2021}+{ }^{2022} C_2(1999)^{2022}(24)+\ldots+\ldots\right) \\\\ & \Rightarrow S_1 \text { is divisible by } 24 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { Now, } S_2: 13(13)^n-12 n-13 \\\\ & \text { Here, } 13^n=(1+12)^n \\\\ & \quad=1+{ }^n C_1 12+{ }^n C_2(12)^2+{ }^n C_3(12)^3 \\\\ & \begin{aligned} \therefore S_2: & 13\left(1+{ }^n C_1(12)+{ }^n C_2(12)^2+{ }^n C_3(12)^3+\ldots\right)-12 n-13 \\\\ \quad= & 13+156 n+13\left({ }^n C_2(12)^2+{ }^n C_3(12)^3+\ldots\right)-12 n-13 \\\\ \quad= & 144 \times 13\left({ }^n C_2+{ }^n C_3(12)+\ldots\right) \end{aligned} \end{aligned} $$ <br/><br/>$\Rightarrow S_2$ is divisible by 144 for infinitely many $n \in N$