Prove that $4^n$ is not divisible by 3.

Question

How can one prove that $4^n$ is not divisible by 3, for any $n \ge 0$?

One way I found is to proof that $4^n - 1$ is always divisible by 3 (as demonstrated in a question here), thus $4^n$ could never be divisible by 3.

Can you suggest a better way to prove this?

Thanks!

Answer 1

If $3$ divides $4^n=2^{2n}$ it would appear in the latter's factorisation into primes.

Answer 2

Use the fact that if a prime divides a product, then it must divide a single term in the product.

Answer 3

$$4=1\mod 3$$ and therefore $$4^n=1\mod 3$$ That's it.

Answer 4

Notice that $4^n=(2^2)^n=2^{2n}$ and $3$ is not a factor of $2^{2n}$.

Answer 5

The prime factors of $4^n$ is the multiset $\{(2, 2n)\}$ and $3 \not \in \{(2, 2n)\}$.