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 $k$.
$a^2 \mid 3b^2$
$a^2 \cdot k = 3b^2$
Where should I go after this step? Is this correct?
Let $t = a^2 \cdot k$ because the product of integers are integers.
$t \mid 3b^2$