2. Find all triples ( $p, q, r$ ) of prime numbers for which

$$
p^{2}-q r=2500
$$
2. Obviously, $p>3$ (otherwise the left side would be smaller than the right), so since $p$ is a prime number, it follows that $p \equiv \pm 1(\bmod 3)$. From there we have $p^{2} \equiv 1(\bmod 3)$. We also have $2500 \equiv 1(\bmod 3)$, so if we transform the given equation into the form

$$
p^{2}-2500=q r
$$

the left side is divisible by 3, so the right side must be as well. Since $q$ and $r$ are prime numbers, one of them must be 3. Without loss of generality, let's assume it is $q$. Note that $2500=50^{2}$, so the left side can be factored as a difference of squares, leaving us with $(p-50)(p+50)=q r=3 r$. Since $r$ is a prime number, we have only the following two possibilities.

- $p-50=3, p+50=r:$ It follows that $p=53$ and $r=103$, which are prime numbers, so we have one solution.
- $p-50=1, p+50=3 r$ : It follows that $p=51$, which is not a prime number, so there is no solution here.

Considering that $q$ and $r$ can switch roles, there are a total of two triples that satisfy the conditions of the problem: $(p, q, r) \in\{(53,3,103),(53,103,3)\}$.