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

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SUMMARY

The discussion centers on proving the statement that for all positive integers k, m, and n, if k divides the product mn (k|mn), then k must also divide either 4m or 4n (k|4m or k|4n). Participants attempted to manipulate the definitions of divisibility but encountered challenges in finding a conclusive proof. The example provided, using k = 21, m = 3, and n = 7, illustrates a case where the statement does not hold, indicating that the original assertion may be incorrect.

PREREQUISITES
  • Understanding of integer divisibility and notation (e.g., k|mn)
  • Familiarity with basic algebraic manipulation
  • Knowledge of positive integers and their properties
  • Experience with constructing mathematical proofs
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  • Study the properties of divisibility in number theory
  • Learn about counterexamples in mathematical proofs
  • Explore the concept of prime factorization and its implications
  • Investigate related theorems in integer arithmetic
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Mathematics students, educators, and anyone interested in number theory or mathematical proofs, particularly those dealing with divisibility concepts.

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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