1. (2 points) Solve the equation $p^{3}-q^{3}=11 r$, where $p, q, r-$ are prime numbers.
Solution: If all three unknown numbers are odd or exactly one of them is odd, then the right and left sides have different parity. The case where all three numbers are equal to two also does not fit. Therefore, among our numbers, there is exactly one two. Moreover, $p>q$, so the two is either $r$ or $q$.

Let $r=2$. Then $p^{3}-q^{3}=(p-q)\left(p^{2}+p q+q^{2}\right)=2 \cdot 11$. The second factor is clearly greater than the first and neither is equal to 1, so $p=q+2$ and $3 q^{2}+6 q+4=11$. This quadratic equation in $q$ has no natural solutions.

The second case: $q=2$. Then $p^{3}-2^{3}=(p-2)\left(p^{2}+2 p+4\right)=11 r$. Clearly, $p=3$ does not work, so one of the factors in $(p-2)\left(p^{2}+2 p+4\right)$ is 11, and the other is $r$. The quadratic equation $p^{2}+2 p+4=11$ has no natural solutions, so the only remaining option is $p-2=11$, i.e., $p=13$. From this, we get $r=199$.