Prove that if $a$ divides $b^2$ then $a$ divides $b$ I'm generalizing this, but in my proof I just need to explain why if $2$ divides $b^2$, then $2$ divides $b$. 
I could find the answer for if $a^2$ divides $b^2$ then $a$ divides $b$, but not for this question. I appreciate any help!
 A: In general, it is not true that if $a$ divides $b^2$, then $a$ divides $b$. For example, $4$ divides $2^2$, but $4$ does not divide $2$.

However, in your case, when $a=2$, it is true.
If $b$ is odd, then $b^2$ is odd so $2$ does not divide $b^2$ - this is not true. Therefore, $b$ is not odd, so $b$ is even. Therefore, $b$ is divisible by $2$.

It is true even more in general, since, if $a$ is prime and $a$ divides $b^2$, then $a$ divides $b$. This is most simply shown using the fundamental theorem of arithmetic.
More generally, this is true if and only if $a$ is not divisible by any square of a prime number. For example, it is true for $a=6$, but it is not true for $a=4$. 
Outline of proof:
Part 1:: If $a$ is divisible by a square of a prime number, then we cannot conclude from $a|b^2$ that $a|b$.
Any time $a$ is divisible by a square of a prime number, $a=k\cdot p^n$ where $n\geq 2$ and $p$ does not divide $k$, you can see that $a$ divides $(k\cdot p^{n-1})^2$, but $a$ does not divide $k\cdot p^{n-1}$.
Part 2: If $a$ is not divisible by a square of a prime number, then we can conclude from $a|b^2$ that $a|b$.
On the other hand, if $a$ is not divisible by a square of a prime number, then $a=p_1p_2\cdots p_n$ where $p_i$ is prime for all $p$. Then, assuming $a$ divides $b^2$, you can conclude that $p_i$ divides $b^2$, and therefore $p_i$ divides $b$ (from prime factorization) for all $i$. From this, you can conclude that $a$ divides $b$.
A: $2$ is a prime number. The idea at play is that if $p$ is prime and if $p$ divides $b^2$ for some integer $b$ then $p$ divides $b$. 
If $p$ divides $b^2 = b \cdot b$ then either $p$ divides $b$ or $p$ divides $b$ (Euclid’s Lemma). So, it follows that $p$ divides $b$.
This doesn’t generalize to all composite integers (see the comments).
A: The general claim you have made that $$a|b^2\implies a|b$$ is not true
As mentioned in the comments, taking the simple counterexample $a=9$, $b=3$
The statement $$2|b^2\implies 2|b$$
is true
Consider the cases: 
Suppose $b$ is even. Then we can write $b=2k\implies b^2=4k^2=2(2k^2)\implies b^2\quad\text{is even}$
Suppose $b$ is odd. Then we can write $b=2k+1\implies b^2 =(2k+1)^2=4k^2+4k+1=2(k^2+2k)+1\implies b^2\quad\text{is odd}$
Then by 'exhaustion' of the possible cases, we have shown that $2|b^2\implies 2|b$
