4.29. For which integers $n$ is the number $20^{n}+16^{n}-3^{n}-1$ divisible by 323?
4.29. Answer: for even n. First of all, 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.