If n is an even number, show that one of the two values n, n+2 must always be a multiple of 4, regardless of the value of n.

Question

I'm currently working through this question and am having some trouble finishing it.

So far I have done the following

For n or n + 2 to be a multiple of 4

n = 4c

n + 2 = 4c

Where c is any integer

Since n is even, n = 2k

So

2k = 4c

2k + 2 = 4c

Where do I go from here to prove that n must be a multiple of 4 in one of these cases for any even value of n?

Answer 1

Here's a hint: if $n$ is even, then $n=2k$. $k$ is either even (say $k=2r$) or $k$ is odd (say $k=2r+1$). Substitute and proceed from there.

Answer 2

If $n$ is even then $n=2k$ for some $k\in\mathbb{N}$. Then $n+2=2k+2=2(k+1)$. Now one of $k, k+1$ must be even as we cannot have two consecutive odd numbers, so one of $n, n+2$ must be divisible by $4$.

Answer 3

$n$ even;

1) If $4|n$ we are done.

2)Assume $4\not | n$, then

$n=2k$ , where $k$ is odd(Why?): $k=2l+1$;

$n=2(2l+1);$

Then

$n+2=2 (2l+1)+2= 4l+4=4(l+1)$.

Hence $4| (n+2)$.