In base $b,$ the square of the number $17_b$ is $211_b$. Find the base $b$.
\(b + 7 = 2b^2+b+1\)
\(2b^2 = 6\)
\(b^2 = 3\)
\(\mathbf{b = \sqrt{3}}\)
But since sqrt(3) < 7, the base wouldn't work for the left side of the equation. Therefore, there is no solution.