Determine all triples $(p, q, r)$ of prime numbers for which $4 q-1$ is a prime number and

$$
\frac{p+q}{p+r}=r-p
$$

is satisfied.

(Walther Janous)

Answer. There is exactly one solution triple, namely $(p, q, r)=(2,3,3)$.
The equation is given in equivalent form as

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

or

$$
4 q-1=4 r^{2}-\left(4 p^{2}+4 p+1\right)
$$

or

$$
4 q-1=(2 r-2 p-1)(2 r+2 p+1) \text {. }
$$

Since $4 q-1$ is a prime number and $2 r+2 p+1>1$, it must be that for the first factor

$$
2 r-2 p-1=1
$$

which means

$$
r=p+1
$$

The only prime numbers for which such a relationship is possible are $p=2$ and $r=3$. Indeed, for this choice,

$$
(2 r-2 p-1)(2 r+2 p+1)=11
$$

is a prime number, and since $4 q-1=11$, it must also be that $q=3$.

Therefore, the only solution is $(p, q, r)=(2,3,3)$.