Disibility proof

[notSomebody]

I'm at a loss here. I have no idea how to prove this.

For all positive integers, k, m, n if k|mn then k|4m or k|4n.

2. Relevant equations
An integer r is divisible by an integer d if and only iff r=ds where s is some integer and d != 0.

3. The attempt at a solution
I tried rewriting the divisibilities.

k|4m
4m = ks

k|4n
4n = kq

k|mn
mn = ky

but I don't know where to go from here.

[daveb]

Are you sure you wrote it right? Take k = 21, m = 3, and n = 7 - doesn't work.

[notSomebody]

Now I look foolish. I tried a couple of arrangements of numbers and it worked out, so I assumed it to be true. Thanks.