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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?

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  • $\begingroup$ Suppose that $n$ is even. Then $n=2k$ for some integer $k$. Now... $k$, being an integer must either be even or odd. In the case that $k$ is even, then $k$ can be rewritten as $k=2\ell$ in which case $n=2k=2(2\ell)=\cdots$. Meanwhile in the case that $k$ is odd then $k$ can be rewritten as $k=2\ell + 1$ in which case $n=2k=2(\cdots)=\cdots$. In the each case, show that either $n$ or $n+2$ is a multiple of $4$. $\endgroup$
    – JMoravitz
    Mar 7, 2019 at 20:29
  • $\begingroup$ Another way: $\,\ 8\cdot 3\,{\Large {n+2\choose 4}} = (n\!+\!2)\,(\overbrace{n\!+\!1}^{\large \rm odd})\,n\,(\overbrace{n\!-\!1}^{\rm\large odd})\,\Rightarrow\, 4\mid n\!+\!2\ $ or $\ 4\mid n\ \ $ $\endgroup$
    – Bill Dubuque
    Mar 7, 2019 at 20:59

3 Answers 3

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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.

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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 $4$.

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$n$ 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 $4| (n+2)$.

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