[Number Theory]
Problem: Compute the smallest positive integer base $b$ for which $16_{b}$ is prime and $97_{b}$ is a perfect square.

Solution: Because 9 is used as a digit, $b \geq 10$. The conditions require that $b+6$ be prime and $9 b+7$ be a perfect square. The numbers modulo 9 whose squares are congruent to 7 modulo 9 are 4 and 5. So $9 b+7=(9 k+4)^{2}$ or $(9 k+5)^{2}$ for some integer $k$. Finally, $b$ must be odd (otherwise $b+6$ is even), so $9 b+7$ must be even, which means that for any particular value of $k$, only one of $9 k+4$ and $9 k+5$ is possible. Taking these considerations together, $k=0$ is too small. Using $k=1$ makes $9 k+4$ odd, and while $(9 \cdot 1+5)^{2}=196=9 \cdot 21+7$ is even, because $21+6=27$ is composite, $b \neq 21$. Using $k=2$ makes $9 k+4$ even, yielding $22^{2}=484=9 \cdot 53+7$, and $53+6=59$ is prime. Thus $b=\mathbf{5 3}$, and $53+6=59$ is prime while $9 \cdot 53+7=484=22^{2}$.

Final answer: 53