- First part of the task is just to show that $(4^n-1)$ actually
is divisible by 3 for n=1,2,3,4. No problem. Second step: is to
show that $(4^n -1) = (2^n-1)(2^n+1)$ No problem, just algebra.
Third step is to explain that $(2^n-1)$,$2^n$,$(2^n+1)$ is three
consecutive numbers. And that only one is divisible by three.
$2^n$ has two as a factor and is not divisible by 3. It can't be
even. That leaves $(2^n-1)$ and $(2^n+1)$. The one with 3 as a
factor seems to be random dependent on n. Fourth step, tie it
all together. Now, I'm not sure if I'm suppose to dig deeper
into which of them is actually divisible by 3. Or has 3 as a
factor. But knowing that one of them at the time (dependent on
n) has indeed 3 as a factor implies that, in respect to the
second step, 3 is a factor of $(4^n -1)$. Is this all there is
to it? Keep in mind that this is a really beginners proof task,
not very formal. It's slightly funny going back to high school
math after being scared to death by logic and descrete math at
university-level