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I understand the problem very well. I just don't how to go at it.

Prove or Disprove: For all , , ∈ ℤ+, if |c, then | or |.

user3233845
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    $12|3\times4=12$, but $12\not|3,12\not|4$. –  Mar 23 '16 at 18:00
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    How to go at it: Write down three integers $a,b,c$. Check whether $a|bc$. If it does, check whether $a|b$ or $a|c$. Repeat with three other integers. Repeat as often as needed until you begin to perceive a pattern. Test the pattern with more repetitions. Try to refine the pattern. Try to find a counterexample. Or try to find a proof. – Lee Mosher Mar 23 '16 at 18:02
  • If a divides b times c does that mean a has to completely divide b or completely divide c? What if part of a divides b and the rest of b divides c but the whole of a divides neither b nor c? Is that possible? What if $k|b$ and $j|c$ and so $a= jk$ and $jk|bc$. Does it follow that $jk|b$ or $jk|c$? – fleablood Mar 23 '16 at 18:08

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Counter-example: $$\large8|6\times 4$$ but $$\large8\not | \, \,6$$ nor $$\large8\not | \, \,4$$

SchrodingersCat
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The trick is to realize that maybe $a$ can be factored to $a = jk$. Then it's possible that $j|b$ and $k|c$ but those two statements are "unrelated". It follows that $jk|bc$ but there is no "relation" between $k$ and $b$ or between $j$ and $c$ so there is no reason to believe either $jk|b$ or $jk|c$.

A counter example is easy to make: Pick any two numbers where one divide the other. Say $2|6$. Then take another pair that are relative prime to both of these were one divides the other. Say $5|35$. Then obviously $2*5|6*35$ but equally obviously $2*5\not \mid 6$ (because $5 \not \mid 6$) and $2*5 \not \mid 35$ (because $2 \not \mid 35$).

(They don't have to be relatively prime. You just need the factors of $a$ "distributed" among $b$ and $c$. If $a = 4 = 2^2$ you can have $2|b$ but $4\not \mid b$ [say $b= 6$ and $2|c$ but $4\not \mid b$ [say $c = 10$]. Then you get $4|60$ but $4 \not \mid 6$ and $4\not \mid 10$.)

But what if $a$ is prime? Then this is true. $a$ can not factor so if $bc$ is a multiple of $a$ either $b$ is, or $c$ is, or both are.

fleablood
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