For all integers of n, $n$ is divisible by $2 \iff n^4$ is divisible by $2$

Question

Provide the proof: $\forall n \in \mathbb{Z}$, $n$ is divisible by $2 > \iff n^4$ is divisible by $2$.

Just curious on how the proof of this statement would look like.

Answer 1

A number $n$ is divisible by two if and only if $n$ is even $\ldots$ equivalently, $n$ is even if and only if $n=2k$ for some $k$.

Now suppose that $n$ is even then $n=2k$ for some $k$ , so

$$n^4 = (2k)^4 = 2^4k^4 = 2\cdot2^3\cdot k^4 = 2\cdot K$$

where $K=2^3\cdot k^4$. Therefore, $n^4$ is even and hence divisible by two. Conversely, suppose $n^4$ is divisible by $2$ but that $n$ were odd. Then repeating the previous we get that $n=2k+1$ for some $k$ and hence that $$n^4 = (2k+1)^4 = 16 k^4 + 32 k^3 + 24 k^2 + 8 k + 1 = 2K+1$$

where $K=8k^4 + 16k^3+24k^2+4k$, but then $n^4$ would be odd and hence not divisible by two (a contradiction).

Notice: This second half of the proof is essentially proof by contraposition but I sort of artificially made it a proof by contradiction (because usually people new to proofs find these easier to follow).

Answer 2

One approach is to prove a more general result, of which your theorem is just a specific instance. That is, let's prove that in general:

$\forall n \in \mathbb{Z}, \forall p \in \mathbb{N}^+: \ \ n$ is divisible by $2 \iff n^p$ is divisible by $2$

This we can prove by induction on $p$:

Base:

$p=1$.

We have to show that:

$$n \text{ is divisible by } 2 \iff n^1 \text{ is divisible by }2$$

Well, that is trivially true, since $n^1=n$

Step:

Thye inductive hypothesis is that for some $k \in \mathbb{N}^+$, we have that:

$$n \text{ is divisible by } 2 \iff n^k \text{ is divisible by }2$$

Now we need to show that:

$$n \text{ is divisible by } 2 \iff n^{k+1} \text{ is divisible by }2$$

We would prove this if we can prove the following two things:

$$n \text{ is divisible by } 2 \Rightarrow n^{k+1} \text{ is divisible by }2$$

and

$$n \text{ is not divisible by } 2 \Rightarrow n^{k+1} \text{ is not divisible by }2$$

The first half: If $n$ is divisible by $2$, then by inductive hypothesis $n^k$ is divisible by $2$ as well, and hence $n$ and $n^k$ are both even. But the product of two even numbers is even, and hence $n^{k+1}=n \cdot n^k$ is even as well

Second half: If $n$ is not divisible by $2$, then by inductive hypothesis $n^k$ is not divisible by $2$ either. Hence, they are both odd. But the product of two odd numbers is odd, and hence $n^{k+1}=n \cdot n^k$ is odd as well, and thus not divisible by $2$