Suppose $(b_a)^2=71_a$, where $a$ and $b$ represent two distinct digits. If $b=a-1$, find $a$.
To solve this problem, I need to figure out what base $a$ the numbers are in, and what digit $b$ is. I know that $b=a-1$, so I can substitute that into the equation and get $(a-1)_a^2=71_a$. I also know that $(a-1)_a$ means $a-1$ in base $a$, which is equivalent to $a-1$ in base 10, since $a-1$ is just one digit. Similarly, $71_a$ means $7a+1$ in base 10, since $7$ is the coefficient of $a$ and $1$ is the constant term. So I can rewrite the equation as $(a-1)^2=7a+1$ in base 10. Expanding the left side, I get $a^2-2a+1=7a+1$. Subtracting $1$ from both sides, I get $a^2-2a=7a$.
[*: Adding $2a$ to both sides, I get $a^2=9a$. Dividing both sides by $a$, I get $a=9$, assuming $a\neq 0$. So the base is 9, and $b=a-1=8$. To check my answer, I can plug in $a=9$ and $b=8$ into the original equation and see if it works. I get $(8_9)^2=71_9$, which means $8^2=7\cdot 9+1$ in base 10. Simplifying, I get $64=64$, which is true. So $a=9$ is the correct answer. 