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Prove or disprove if $a\mid b$ and $a^2\mid c$ then $b\mid c$.

My idea:

Since $a\mid b$ then there exists a $k$ such that $ak=b$.

Since $a^2\mid c$ then there is a $k_1$ such that $a^2k_1=c$.

I did't get any idea from here is that statement is true or false.

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  • $\begingroup$ if you can conclude from there, then search instead for a counter example. $\endgroup$ – zwim Oct 3 '17 at 4:18
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This is false. For a counterexample take $a=1$, $b=2$, and $c=3$.

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The statement is false. Hint: what happens when $a=0$?

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Okay, everyone is just telling you the answer but let's see how you can actually prove that. Keep in mind that if you assume one way and try and prove it you'll either get a contradiction or no issues - so it really doesn't matter as long as you structure your argument correctly.

Consider that if $a | b$ then we can write $ai = b$ for some integer $i$ - great, you got that. Now, let's say the same thing for the other statements.

$a^2 | c \Rightarrow a^2j = c$ and $b | c \Rightarrow bk = c$

now, try using these things to show that it either does or doesn't make sense:

$ai = b\\ a^2j = c\\ bk = c$

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