How can I show that $4^{1536} - 9^{4824}$ can be divided by $35$ without remainder? How can I show that $4^{1536} - 9^{4824}$ can be divided by $35$ without remainder?
I'm not even sure how to begin solving this, any hints are welcomed!
$$(4^{1536} - 9^{4824}) \pmod{35} = 0$$

Since $35=7*5$ and $5,7$ are prime, it suffices to show that that is divisible by both $5$ and $7$. To do this, note that $4\equiv 9\equiv -1$ mod $5$, so 
$$
4^{1536}-9^{4824}\equiv (-1)^{1536}-(-1)^{4824}\equiv 1-1\equiv 0\mod 5.
$$
To show divisibility by $7$ requires a bit more work; try to show that successive powers of an element eventually cycle mod $7$, and then you will be able to get a nice form for them.
