In base $b,$ the square of the number $17_b$ is $211_b$. Find the base $b$.

Answer 1

Let $a$ be the digit in the units place of $17_b.$ Then, $17_b=a\cdot b^0=a.$

Since the square of $17_b$ is $211_b,$ we have
\[(10a+7)^2=2b^2+b+1.\]Expanding, we get
\[100a^2+140a+49=2b^2+b+1.\]Since $a$ is the units digit of $17_b,$ $0\le a<b.$ Thus, we must have $a=1,$ which implies $b=7.$

Therefore, the base $b$ is $\boxed{7}.$