A sum of powers of $2$ or $4$ that is or isn't divisible by $3$

Question

Let $n\in\mathbb{Z^+}$ (a positive integer), and define $E(n)=2^{2n}$ and $O(n)=2^{2n+1}$.

Alternatively we can define $E$ to be $4^n$ and $O$ to be $2\cdot4^n\\$.

Let $a$ and $b$ be some arbitrary positive integer where $a>b$. I empirically found out that $$\frac{E(a) - E(b)}{3}$$ after testing a bunch of numbers. Hence, $E(a)-E(b)$ is always divisible by $3$.

The same applied to $O(a)-O(b)$.

How do I mathematically prove that $E(a) - E(b)$ and $O(a) - O(b)$ is always divisible by $3$, but $E(a) - O(b)$ is not divisible by $3$?

Im not sure if these are common theorem, I could not find online. I have a hard time figuring out where to start and proving techniques since Im didnt really understood number theory proving yet.

Answer 1

$4 \equiv 1 \mod 3$, so $4^n \equiv 1^n \equiv 1 \mod 3$, and $2 \cdot 4^n \equiv 2 \mod 3$.

Answer 2

Assume that $a>b$.

$$E(a)-E(b)=4^a-4^b=4^b(4^{a-b}-1)$$

$$=4^b.(4-1)(1+4+4^2+...+4^{a-b-1})$$ $$=3.4^b(1+4+4^2+...+4^{a-b-1})$$ $$\equiv 0 \mod 3$$