Pluggin in numbers If $(a + c)(a − c) = 0$, which of the following must be true?


*

*$a = 0$,

*$c = 0$,

*$a = −c$,

*$a = c$,

*$a2 = c2$


The answer states: Try plugging in $a = 2$ and $c = 2$. This eliminates (A), (B), and (C). Now try $a = 2$ and $c = −2$. This eliminates (D). Only (E) remains. 
I don't understand how the plug in numbers were chosen. Could anyone help explain this?
 A: By choosing $a = 2$, $c = 2$, the author demonstrates that the equation can be satisfied when $a \neq 0$ since 
$$(a + c)(a - c) = (2 + 2)(2 - 2) = 4 \cdot 0 = 0$$
Hence, $a = 2$, $c = 2$, is a counterexample to the claim that $a = 0$ must be true.  There are others.  For instance, $a = 1$, $c = 1$ or $a = 1$, $c= -1$. However, only one counterexample is required to demonstrate that a statement is false.
What the author is stating is that $a = 2$, $c = 2$ is also a counterexample to the claim that $b = 0$ must be true or that $a = -c$ must be true for the equation $(a + c)(a - c) = 0$ to be satisfied.
Similarly, if $a = 2$ and $c = -2$, then 
$$(a + c)(a - c) = [2 + (-2)][2 - (-2)] = 0 \cdot 4 = 0$$
so $a = 2$, $c = -2$ is a counterexample to the claim that $a = c$ must be true for the equation $(a + c)(a - c) = 0$ to be satisfied.  Again, there are other counterexamples such as $a = 1$, $c = -1$.  
As for the remaining choice, since $(a + c)(a - c) = a^2 - c^2$, $$(a + c)(a - c) = 0 \Rightarrow a^2 - c^2 = 0 \Rightarrow a^2 = c^2$$
A: Another way to come to the same conclusion is to multiply the expression out: $$ (a+c)(a-c) = a^2 - c^2 $$ so if $(a+c)(a-c) = 0$ then you also have $a^2 - c^2 = 0$.  Adding $c^2$ to each side shows (E) is a correct answer.  
This allows you to sidestep the written answer's ad hoc choices of values which they use to eliminate the first 4 options.
