Let a and b be non-zero integers. If both a^2 and b^2 have at most two digits, then what is the greatest possible difference between a^2 and b^2?
Since $a^2$ and $b^2$ each have at most two digits, $-10\leq a,\,b\leq 10.$ The equation $a^2-b^2=(a+b)(a-b)$ states that the difference between two squares is expressible as a difference of two products of integers. Note that $a+b$ and $a-b$ have the same parity for integers $a,$ $b.$
The maximum possible difference of two numbers of the same parity is achieved when the two numbers have a maximum possible distance between them. Because $-10\leq a,\,b\leq 10,$ the greatest possible gap between two numbers is attained when we choose $10$ and $-10.$ Thus, $a+b=10+(-10)=0$ and $a-b=-10-10=-20,$ resulting in the greatest possible difference: \begin{align*}
a^2-b^2&=(a+b)(a-b)\\
&=(0)(-20)\\
&=\boxed{0}.
\end{align*}
NO
Apologies, I made an error in my response. Let me correct it.
Since $a^2$ and $b^2$ each have at most two digits, we have $10 \leq a^2 \leq 99$ and $10 \leq b^2 \leq 99$. So $4 \leq a \leq 9$ and $a = -b$ (since $b$ is non-zero).
Now, we want to maximize $a^2 - b^2$ which can be written as $(a + b)(a - b)$. We want to maximize $a + b$ since $a - b$ is always negative.
Since $a = -b$, we have $a + b = -b + b = 0$. So the greatest possible difference between $a^2$ and $b^2$ is $\boxed{0}$.
Try again and give a different answer.
Apologies for the confusion. Let's try this again.
Since $a^2$ and $b^2$ each have at most two digits, we have $10 \leq a^2 \leq 99$ and $10 \leq b^2 \leq 99$. So $4 \leq a \leq 9$ and $4 \leq -b \leq 9$. This means $-9 \leq b \leq -4$.
To maximize the difference between $a^2$ and $b^2$, we want to choose the largest possible value for $a$ and the smallest possible value for $b$.
Let $a = 9$ and $b = -9$. Then $a^2 - b^2 = 9^2 - (-9)^2 = 81 - 81 = \boxed{0}$.