Suppose $(b_a)^2=71_a$, where $a$ and $b$ represent two distinct digits. If $b=a-1$, find $a$.

We can start by replacing b with (a-1).

$$\begin{align*} (a-1)^2\cdot a^0&=7\cdot a^1+1\cdot a^0 \\\Rightarrow\qquad a^2-2a+1&=7a+1 \\\Rightarrow\qquad a^2-9a&=0 \\\Rightarrow\qquad a(a-9)&=0 \end{align*}$$

So we can have 0 or 9.

But there is no base 0, so must be base 9. 

thanks

Oh yay! My LaTeX finally worked for once.