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Hello I'm having trouble finding a proof for this problem. I have the following proof so far....

Suppose $a$ and $b$ are particularly chosen integers such that $a\mid b$.

By definition of divisibility $b= a\cdot k$ for some integer $k$.

Then,

$a^2 \mid 3b^2$

$a^2 \cdot k = 3b^2$

Where should I go after this step? Is this correct?

Let $t = a^2 \cdot k$ because the product of integers are integers.

Therefore

$t \mid 3b^2$

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    $\begingroup$ Since $b=ak$, we have $3b^2=3(ak)^2=a^2(3k^2) \ldots$ $\endgroup$ – Francesco Polizzi Jul 2 '18 at 21:33
  • $\begingroup$ I can see how 3b² = 3(ak)² but cant see how that is equivalent to a²(3k²) . Could you explain a bit? And what would be the correct category to ask this question? $\endgroup$ – Jlee Jul 2 '18 at 21:47
  • $\begingroup$ Maybe the answers posted here can help you: Show that for all integers $a$ and $b$ if $a\mid b$ then $a^2\mid b^2$ $\endgroup$ – Martin Sleziak Jul 3 '18 at 11:00
  • $\begingroup$ @Jlee As Francesco Polizzi has explained above, just square both sides of $b=ak$ and rearrange to get the desired conclusion. $\endgroup$ – Allawonder Jul 3 '18 at 11:03
  • $\begingroup$ "I can see how 3b² = 3(ak)² but cant see how that is equivalent to a²(3k²) " That's just rewriting. Commutativity and associativtiy. $3(ak)^2 = 2\times a\times k\times a\times k = a\times a\times 3\times k\times k = a^2(3k^2)$. This was not meant to be tricky. $\endgroup$ – fleablood Jul 4 '18 at 4:00
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You have things backward. Your first sentence is right, that $b=ak$. But then you leap to the very end and state that "Then $a^2 \mid 3b^2.$" But that's your conclusion. It should be the last sentence of your proof.

You need to start with $b=ak$ and work your way to $3b^2 = a^2n$ (for some integer $n$) and then you can conclude that $a^2 \mid 3b^2.$

You might write: Well, $b=ak$, so $b^2 = a^2k^2$, so $3b^2 = 3a^2k^2 = a^2(3k^2).$ Since $3k^2$ is an integer,$\ldots$

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It's even easier than that. if $a \mid b$, then perforce $a^2 \mid b^2$ and hence $a^2 \mid nb^2$ for all integers $n$, including $3$.

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