Problem 903

It is

$$k=46^{2 n}-12^{2 n}=2^{2 n}\left(23^{2 n}-6^{2 n}\right)=4^{n}\left[\left(23^{n}\right)^{2}-\left(6^{n}\right)^{2}\right]=4^{n}\left(23^{n}+6^{n}\right)\left(23^{n}-6^{n}\right)$$

Thus, the number $k$ is certainly divisible by 4. Furthermore,

$$23^{n}-6^{n}=(17+6)^{n}-6^{n}=\sum_{i=0}^{n}\binom{n}{i} 17^{i} 6^{n-i}-6^{n}=\sum_{i=1}^{n}\binom{n}{i} 17^{i} 6^{n-i}$$

which means the number $k$ is divisible by 17. Finally,

$$23^{n}-6^{n}=(29-6)^{n}-6^{n}=\sum_{i=0}^{n}\binom{n}{i} 29^{i} 6^{n-i}-6^{n}=\sum_{i=1}^{n}\binom{n}{i} 29^{i} 6^{n-i}$$

which means the number $k$ is divisible by 29. Therefore, $k$ is divisible by $4 \cdot 17 \cdot 29=1972$.