Why is it that $\forall n \in N$, $2^n$ is not divisible by $3$?
I can prove it easily by induction, but I don't understand the intuition of why this is true. Could anyone please supply the intuition? [This has been answered.]
Moreover, I understand that there's also a proof of this fact that uses modular arithmetic. How does this proof go? [Also answered - it follows immediately from the definition of the modulus operation.]
While $\,3\nmid 2^n\,$ follows by FTA = existence & uniqueness of prime factorizations, we can give a much simpler modular arithmetic proof that $\rm \ a\nmid (a\!-\!1)^n =: \color{#0a0}b,\,$ for all integers $\rm\ \color{brown}{a\ne \pm1},\ $namely
$\rm mod\,\ a\!:\,\ \color{#c00}{a\equiv 0}\ \Rightarrow \color{#0a0}b = (\color{#c00}{a}\!-\!1)^n\equiv (\color{#c00}{0}\!-\!1)^n \equiv \pm1 \color{#0a0}{\not\equiv 0}\ \ (else\ \ \color{brown}{a\mid\pm1})\,$ by Congruence Laws
So we have shown $\rm\ \color{#0a0}{b\not\equiv 0}\pmod a,\:$ i.e. $\rm\,\ a\nmid b.\ \ \bf\small QED$
Alternatively, if modular arithmetic is unfamiliar, we can prove it by induction, with the inductive step being $\rm\ a\mid \color{brown}{(a\!-\!1)n}\ \Rightarrow\ a\mid n = an-\color{brown}{(a\!-\!1)n}\ $ [which is a special case of Euclid's Lemma, i.e. $\rm\, (a,m)=1,\ a\mid mn\,\Rightarrow\, a\mid n\,$]. $ $ So $\rm\ a\mid (a\!-\!1)^n \Rightarrow\ a\mid (a\!-\!1)^{n-1} \Rightarrow \cdots \Rightarrow\ a\mid 1.$
Or we can use the Binomial Theorem to write $\rm\, (a-1)^n = (\cdots)a + (-1)^n,\,$ which is essentially equivalent to the first proof using congruence arithmetic.
$2^n$ has a prime factorization containing only $2$s. So it can never have 3 has one of its prime factors by the Unique Factorization Theorem.
I can think of the proof using this.
Suppose for a contradiction that $2^n$ is divisible by 3. Then $$2^n\equiv 0\mod3$$
That means $3$ divides $2^n$. Now $2^n =3k$ for some $k\in \mathbb{Z}$.
Use the first statement I mentioned above to arrive at a contradiction.
$$2^n=\underbrace{2\times2\times\cdots\times2}_{\text{$n$ times}}$$
But for a number to be divisible by $3$ it requires that one of its prime factors is $3$, whereas in the prime factorization of $2^n$ only $2$ appears. Therefore, $2^n$ is not divisible by $3$ for $\forall n\in\mathbb N$.
In the same sense we can generalize this result: If $p$ is a prime number and $n\in\mathbb N$ then $p^n$ cannot have another prime factor other than $p$, it is therefore divisible only by $p$ or $p$ to some power. Example: $5^4$ is not divisible by $2$ nor $3$ for instance but it is divisible by $5$ and $25=5^2$.
If $p$ is prime then $p\mid k\times m$ implies that $p\mid k$ or $p\mid m$. Applying that (eventually more times) on $3\mid 2^n$ leads to $3\mid 2$ wich is not true.
An intuitive way to see it is by writing $2 = 3 - 1$, so we get $$ 2^n = (3 - 1)^n, $$ which when expanded out, has a bunch of terms that are multiples of $3$ and only a single remaining term $(-1)^n$ at the end. So it is a multiple of $3$ plus or minus $1$, and hence not divisible by $3$.
For example, when $n = 2$, we get $$ (3-1)(3-1) = 3 \cdot 3 - 3 - 3 + 1 = \underbrace{(9 - 3 - 3)}_{\text{multiple of } 3} + 1. $$ When $n = 3$, we get $$ (3-1)^3 = \underbrace{3^3 - 3 \cdot 3^2 + 3 \cdot 3}_{\text{multiple of } 3} - 1. $$
And when $n = 4$: $$ \begin{align*} (3-1)^4 &= (3-1)(3-1)(3-1)(3-1) \\ &= 3^4 - 4 \cdot 3^3 + 6 \cdot 3^2 - 4 \cdot 3 + 1 \\ &= \underbrace{(3^3 - 4 \cdot 3^2 + 6 \cdot 3 - 4) \cdot 3}_{\text{multiple of } 3} + 1. \end{align*} $$
