1. If $q$ is a prime number, and $q+1$ is a perfect square, then $q$ is called a $P$-type prime. There are ( ) such $P$-type primes.
(A) 0
(B) 1
(C) 2
(D) infinitely many
- 1. B.

From the given, let $q+1=m^{2}\left(m \in \mathbf{Z}_{+}, m \geqslant 2\right)$. Then $q=m^{2}-1=(m+1)(m-1)$.
Thus, when $m \geqslant 3$, $q$ must be a composite number, which contradicts the problem statement.

Therefore, $m=2$. In this case, $q=3$ meets the condition. Hence, there is only one P-type prime.