Show $ $ $\frac{2n}{n-2} \notin \mathbb{N}$ $ $ for $ $ $n>6$. How do I show $\frac{2n}{n-2} \notin \mathbb{N}$ for $n>6$?

This showed up in an unrelated problem as a condition to validate my solution.  I've tried induction, but I'm not seeing how to prove $ $ $\frac{2k}{k-2} \Rightarrow \frac{2(k+1)}{(k+1)-2}$.
 A: $$\frac{2n}{n-2} = \frac{2n-4+4}{n-2} = 2+\frac{4}{n-2}$$
If $n>6$, $\frac{4}{n-2}<1$, meaning it cannot be an integer.
If $\frac{4}{n-2}$ cannot be an integer, neither can $2+\frac{4}{n-2}$
Therefore, $\frac{2n}{n-2} \notin \mathbb{N}$ for $n>6$
A: This is true for all $n>6$. Suppose not, i.e., $m=\frac{2n}{n-2}$ is an integer $>6$. then for $n$ even say $n=2k$, then $m=2\cdot\frac{2k}{2k-2}=2\frac{k}{k-1}$, but   $k\in \mathbb{N}$ then $\gcd(k,k-1)=1$ for all $k> 3$, which is impossible. For $n$ odd say $n=2k-1$ then $m=2\cdot\frac{2k-1}{2k-1-2}=2\frac{2k-1}{2k-3}$, since (2k-1)-(2k-3)=2, then $\gcd(2k-1,2k-3)=1$, which is also impossible to get an integrer.
A: Suppose $\dfrac{2n}{n-2}=k$ for some integer $k$ and for some $n>6$. We show that $k$ must be less than $n$:
$$n>6 \implies n>4 \implies n-4>0 \implies n(n-4)>0 $$
$$\implies n^2-2n>2n \implies n(n-2)>2n \implies n> \dfrac {2n}{n-2} \implies n>k$$
But $\dfrac{2n}{n-2}=k \implies 2n=nk-2k \implies n(2-k)=-2k \implies \dfrac {2k}{k-2}=n$, which surprisingly has the same form as the original.
Now if $k>4$, we can apply the same long argument as above to conclude $k>n$, which would be a contradiction. Thus $k \le 4$. But plugging in $k=1, 2, 3, 4$ and solving for $n$ shows that $n$ is not an integer greater than $6$, which is a contradiction again.
A: Induction isn't the best way to go here.  Instead, notice $\frac{2n}{n-2} = 2+\frac{4}{n-2}$. When would this be an integer?
