Show that $121^{n}-25^{n}+1900^{n}-(-4)^{n}$ is divisible by 2000. My question comes from the British Mathematical Olympiad (Round 1) paper from 2000:
Show that $121^{n}-25^{n}+1900^{n}-(-4)^{n}$ is divisible by 2000 for any natural $n$.
My immediate idea was to write $121^{n}-25^{n}=(11^{n}
+5^{n})(11^{n}-5^{n})$ but this doesn't seem to help.
My next idea was to use induction; the base case is easy but I can't see how the $k^{th}$ case would imply the $(k+1)^{th}$. Any help would be appreciated.
Factor $2000 = 2^5 \cdot 5^3$

then split $$121^{n}-25^{n}+1900^{n}-(-4)^{n}$$ into


$$9^{n}-9^{n}+12^{n}-12^{n} \equiv 0 \pmod {2^4}$$
$$121^{n}-25^{n}+25^{n}-121^{n} \equiv 0 \pmod {5^3}$$


using Chinese remainder theorem.
Do it by induction. Does it work for $n = 1$? Assume it works for some $n = k$ and show it holds for $n = k + 1$.