2. Find the smallest natural number $m$, for which the expression $148^{n}+m \cdot 141^{n}$ is divisible by 2023 for any odd natural $n$. (16 points)
Solution. $2023=7 \cdot 289$, GCD $(7 ; 289)=1$. Since $n-$ is an odd number, then $148^{n}+m \cdot 141^{n}=(289-141)^{n}+m \cdot 141^{n}=289 l+(m-1) 141^{n}, l \in \square$. Then $(m-1) 141^{n}$ must be divisible by 289. Since 289 and 141 are coprime, then $m-1=289 k, k \in\{0\} \cup \square$. On the other hand $148^{n}+m \cdot 141^{n}=148^{n}+m \cdot(148-7)^{n}=7 p+(m+1) 148^{n}, p \in \square$. Since 7 and 148 are coprime, then $m+1=7 s, s \in \square$. Then $m=289 k+1$ and $m=7 s-1$, and $289 k+1=7 s-1$, $289 k+2=7 s, \quad 287 k+2 k+2=7 s, \quad 2 k+2=7(s-41 k)$. The number $(s-41 k)=2 q, q \in \square$, and $k=7 q-1, k \geq 6$. For $k=6$ we have $m=289 \cdot 6+1=1735$, and $s=248$.

Answer: 1735.