TITLE: Prove that $7^{100} - 3^{100}$ is divisible by $1000$
QUESTION [8 upvotes]: Prove that $7^{100} - 3^{100}$ is divisible by $1000$

Equivalently, we want to show that $$7^{100} = 3^{100} \pmod {1000}$$
I used WolframAlpha (not sure if that's the right way though) and found that $\varphi (250) = 100$.
So by Euler's theorem: $$7^{100} \equiv 7^{\varphi(250)} \equiv 1 \pmod {250} \\ 3^{100} \equiv 3^{\varphi(250)} \equiv 1 \pmod {250}$$
but of course, we want $\pmod {1000}$.
Is that what I'm intended to do in this exercise (how to proceed if so)? Is there a solution without the need to use WolframAlpha?
Thanks!

REPLY [2 votes]: $$(5+2)^{100}-(5-2)^{100} = \sum_{\substack{0\leq k \leq 100\\ k\text{ odd}}}\binom{100}{k} 5^k 2^{101-k}$$
hence the LHS $\pmod{5^3}$ is just $\binom{100}{1}5^1 2^{100}$, i.e. zero. The LHS is also a multiple of $8$ since any odd square is $\equiv 1\pmod{8}$. By the Chinese remainder theorem $7^{100}-3^{100}\equiv 0\pmod{1000}$.