3-ча 1. For which integers \(n\) is the number \(20^{n}+16^{n}-3^{n}-1\) divisible by 323?
Solution 1. Answer: for even \(n\). First, note that \(323=17 \cdot 19\), so a number is divisible by 323 if and only if it is divisible by 17 and 19. The number \(20^{n}-3^{n}\) is divisible by \(20-3=17\). Furthermore, \(16^{n} \equiv(-1)^{n}(\bmod 17)\), so the number \(16^{n}-1\) is divisible by 17 if and only if \(n\) is even. As for divisibility by 19, \(20^{n}-1\) is divisible by \(20-1=19\) for any \(n\), and for \(n=2 m\), the number \(16^{n}-3^{n}\) is divisible by \(16^{2}-3^{2}=13 \cdot 19\), so it is divisible by 19.