Show that for all integers $a$ and $b$ if $a\mid b$ then $a^2\mid b^2$ For all integers $a$ and $b$, If $a\mid b$ then $a^2\mid b^2$ .
So far I have: $b|a \Rightarrow b=ak$ $k$ is an integer, $b^2|a^2 \Rightarrow b^2=a^2q$, $q$ is an integer 
where do I go from here? I was thinking of taking $b=ak$ and multiplying both sides by $b$ but I'm not sure.. 
 A: I don't quite understand your solution, but it looks like you are on the right track. 
If $a \mid b$, then $b = ak$ for some integer $k$. Then $b^2 = k^2 a^2$ so $a^2 \mid b^2$.
A: Hint $\ $ Clear if $\ a=0.\,$ Else $\ a\mid b\ \Rightarrow\ \dfrac{b}a\in\Bbb Z\ \Rightarrow\ \dfrac{b^2}{a^2} = \left(\dfrac{b}a\right)^2\!\in\Bbb Z\ \Rightarrow\ a^2\mid b^2$
i.e. integers are preserved by squaring implies so too are divisibility relations.
A: Here's a hint, whenever you are doing a logic if {}, then{}, start with one half of the then statement. Set it equal to itself. This must be true. Then use the other true things from the if{} to plug into it. 
Hope that makes sense
Read on if you want to know the answer --- ^_^
if $a | b$ is the same as $ak = b$, right!
$a^2 | b^2$ is the same as $a^2 k = b$. We can't assume this part is true because it is in the then part of the if-then, but we can very safely assume $a^2 k = a^2 k$. 
$a^2 k = a^2 k$
$a^2 k = (b^2 / k ^2) * k$
$a^2 k^2 = b^2$
$a^2 k^2 = b^2$
Lemma: if k is an integer, k^2 is an integer
let k^2 = n  (because k^2 is an integer, let's call it a new integer n for simplicity)
$a^2 n | b^2$
$a^2 | b^2$  
Also I reread your original post. Multiplying by b and then substituting for b would work. You had an idea you should have seen it through! When you are unsure you must guess and check your hunches. Only when you are sure, then there is no need to guess and check.
A: You're given that $\;a \mid b\;$, or in other words, $\;\langle \exists l :: l \times a = b \rangle\;$.  (I'm leaving implicit the fact that all variables range over the integers.)  So let us choose such an $\;l\;$, and see what we can make of $\;a^2 \mid b^2\;$:
\begin{align}
& a^2 \mid b^2 \\
\equiv & \qquad \text{"definition of $\;\mid\;$"} \\
& \langle \exists k :: k \times a^2 = b^2 \rangle \\
\equiv & \qquad \text{"use $\;l \times a = b\;$; simplify"} \\
& \langle \exists k :: k \times a^2 = l^2 \times a^2 \rangle \\
\Leftarrow & \qquad \text{"logic: weaken using Leibniz' rule -- to get rid of the $\;a\;$"} \\
& \langle \exists k :: k = l^2 \rangle \\
\equiv & \qquad \text{"$\;l^2\;$ is an integer"} \\
& \text{true} \\
\end{align}
