Suppose $a,b \in Z$. The number $(a-3)b^{2}$ is even if and only if $a$ is odd or $b$ is even. 
*

*Suppose $a,b \in \Bbb Z$. The number $(a-3)b^{2}$ is even if and only if $a$ is odd or $b$ is even.


My plan proof. $(\Rightarrow)$ Assume $(a-3)b^{2}$ is even. 


*

*Assume $a$ is even. Then I will show $b$ is even.

*Assume $b$ is odd. Then, I will show $a$ is odd.


$(\Leftarrow)$. 


*

*Assume $a$ is odd. I will show $(a-3)b^{2}$ is even.

*Assume $b$ is even. I will show $(a-3)b^{2}$ is even.


Can you check my plan proof?
 A: Your proof looks good to me.
I’ve filled it in for you below, following your plan, just in case.
$$(\Rightarrow)$$
Assume $(a-3)b^2$ is even and $a$ is even. Then $a-3$ is odd. So $b^2$ must be even since $mn \in \mathbb{Z}$ is even only if at least one of $m$, $n$ are even. By the same logic, $b$ is even. Now assume $b$ is odd. Then $b^2$ is also odd. So $(a-3)$ must be even since $mn$ is odd only if both $m$ and $n$ are odd. So $a$ must be odd.
$$(\Leftarrow)$$
Assume $a$ is odd. Then $(a-3)$ is even. Since one of the factors is even, $(a-3)b^2$ is even. Assume $b$ is even. Then similarly $(a-3)b^2$ is even.
A: Your proof looks fine to me. In fact, all cases are considered and you've "shown" the bidirectionality of the statement. Why "shown"? 
Because 'Then I will show $b$ is even' (see $\Rightarrow 1)$ isn't really a proof but telling you'll prove it...

Sidenote: Although a case-by-case analysis is almost always a rigorous way to approach a problem, it's often not the nicest. You could for instance take the term modulo $2$: 

$$\begin{align*}(a-3)\cdot b^2\equiv 0\pmod 2&\stackrel{\text{zero product}} \iff  &&a-3\equiv 0\pmod 2 \lor b^2\equiv 0\pmod 2\\
&\;\,\,\iff  &&a\equiv 1\quad\;\,\pmod 2\lor b\;\,\equiv 0\pmod 2
\end{align*}$$ 

