Proving Divisibility of Integers: k|mn, k|4m, k|4n

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The discussion revolves around proving that for all positive integers k, m, and n, if k divides the product mn, then k must also divide either 4m or 4n. The original poster expresses confusion and uncertainty about how to approach the proof, despite attempting to rewrite the divisibility conditions. They provide a counterexample using k = 21, m = 3, and n = 7, which does not support the claim. The conversation highlights the difficulty in validating the statement and the need for a clearer understanding of divisibility rules. Ultimately, the assertion remains unproven and requires further exploration.
notSomebody
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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.

Homework Equations


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

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.
 
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Are you sure you wrote it right? Take k = 21, m = 3, and n = 7 - doesn't work.
 
Now I look foolish. I tried a couple of arrangements of numbers and it worked out, so I assumed it to be true. Thanks.
 

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