What natural numbers $k>2$ satisfy $k-2|2k$?

Question

Find all positive integers $k>2$ such that $k-2|2k$.

I am stuck, I have tried everything. Also, I came up with an answer but I regarding the topics we are seeing currently in class I should be using modular arithmetic.

What I found was that $\lim\limits_{k \to \infty} \frac{2k}{k-2} = 2$. So, I should find the $k$ that causes $\frac{2k}{k-2} = 3$ which is $k = 6$ so $k \le 6$ and plugging in the integer values for $3 \le k \le 6$ I get $k = 3, 4, 6$.

The problem is that I think I should be using modulos and all that and I can't see how to do it that way.

Answer 1

$\frac{2k}{k-2} = 2 + \frac{4}{k-2} \Rightarrow (k-2)|4 \Rightarrow k=3,4,6$.

There's no need to use limits or modules.

Answer 2

First, $k-2|2(k-2) \implies k-2|2k-4$.

Therefore, $k-2|2k \implies k-2\leq4 \implies k \leq6$.

And for that, you've already got the solutions of $k=3,4,6$.

Answer 3

You can do it modular, but it is it basically the same approach: $2k \equiv 0 \mod k-2$, so $4 \equiv 0 \mod k-2$, so $(k-2)\mid 4$. This again gives $k=3,4,6$.