Find all prime numbers $p, q, r$ (not necessarily positive) such that

$$
\frac{1}{p-q-r}=\frac{1}{q}+\frac{1}{r}
$$
If the equality holds for $p, q$, and $r$ (not necessarily positive) prime numbers, then $p-q-r \neq 0 (q \neq 0, r \neq 0)$; multiplying both sides by $q r(p-q-r)$:

$$
r q=(p-q-r) \cdot r+(p-q-r) q, \quad \text { i.e., } r q=(p-q-r)(r+q)
$$

Since $r$ and $q$ are primes, the product $r q$ can have 8 divisors: $1, -1, r, q, -r, -q, r q, -r q$, so $p-q-r$ and $r+q$ can only be one of these. Let's examine all cases:
I. $p-q-r=1$. Then $r+q=r q$, or $0=r q-r-q$, from which $1=(q-1)(r-1)$, so $q$ and $r$ are either 0 (which would not be primes), or $q=r=2$, from which $p=5$.

II. $p-q-r=-1$. Then $r+q=-r q, r q+r+q=0, (r+1)(q+1)=1$, so $r$ and $q$ are either 0 (which would not be primes), or $q=r=-2$, from which $p=-5$.

III. $p-q-r=r$. Then $r+q=q$, from which $r=0$, which is not a prime.

IV. $p-q-r=q$. Then $r+q=r$, from which $q=0$, which is not a prime.
V. $p-q-r=-r$. Then $r+q=-q, r=-2 q$, which is not possible if both $r$ and $q$ are primes.

VI. $p-q-r=-q$. Then $r+q=-r, q=-2 r$, which is also not possible if both $r$ and $q$ are primes.

VII. $p-q-r=r q$. Then $r+q=1$, which is only possible for the primes $r$ and $q$ if $r=-2, q=3$ or $r=3, q=-2$. In both cases, $p-1=-6$, so $p=-5$.

VIII. $p-q-r=-r q$. Then $r+q=-1$, which is only possible for the primes $r$ and $q$ if $r=-3, q=2$ or $r=2, q=-3$. In both cases, $p+1=6$, so $p=5$.

It can be seen that we have examined all cases, and neither $p-q-r$ nor $r+q$ can take on other values, so the problem has the following 6 solutions:

| $p$ | $q$ | $r$ |
| ---: | ---: | ---: |
| 5 | 2 | 2 |
| -5 | -2 | -2 |
| -5 | 3 | -2 |
| -5 | -2 | 3 |
| 5 | 2 | -3 |
| 5 | -3 | 2 |

Ilona Spancer (Nagykanizsa, Batthyány L. Gymnasium, 9th grade)

Note. Similarly, the problem can also be solved by case analysis if we examine when $\frac{r q}{r+q}$ can be an integer, as demonstrated in the solution by Gergely Nagy (Veszprém, Lovassy L. Gymnasium, 10th grade).