How many bases $b \ge 2$ are there such that $100_b + 1_b$ is prime?

Answer 1

Let's first convert $100_b$ and $1_b$ to base 10 in terms of $b$:

$100_b = b^2 + 0b^1 + 0b^0 = b^2$

$1_b = b^0 = 1$

Therefore, $100_b + 1_b = b^2 + 1$. For this sum to be prime, $b^2 + 1$ must be a prime number.

However, we know that for any integer $n \geq 2$, $n^2 + 1$ is always composite because $(n + 1) \mid (n^2 + 1)$. Therefore, there are \(\boxed{0}\) bases $b \ge 2$ such that $100_b + 1_b$ is prime.