## Task 14/86

Let $N$ be the set of natural numbers and $P$ be the set of prime numbers. We are looking for all solution triples $(x ; y ; z)$ with $x ; y \in P ; z \in N$ of the equation

$$
\frac{x+y}{x-y}=z
$$
Due to $z=\frac{x+y}{x-y}=1+\frac{2 y}{x-y} \in N$, it follows that $\frac{2 y}{x-y} \in N$. Consequently, $x-y$ is a divisor of $2 y$. Because of the prime property of $y$, there are only 4 possibilities:

1) $x-y=1, x=y+1$. This means $x$ and $y$ are consecutive numbers; due to the prime property of $x$ and $y$, this is only possible for $y=2, x=3$. It follows that $z=5 \in P \subset N$.
2) $x-y=2, x=y+2$. Then $z=\frac{x+y}{x-y}=\frac{2 y+2}{2}=y+1$. Solution triples arise for all prime twin pairs, where $y$ is the smaller of the two primes and $z$ is the natural number in between.
3) $x-y=y, x=2 y$. This means $x \notin P$, so this case is impossible.
4) $x-y=2 y, x=3 y$. This corresponds to case 3.

Addition: If we assume $x \in N$ instead of $x \in P$, then cases 3 and 4 also yield solutions:

3) $(2 y ; y ; 3)$ with $y \in P$.
4) $(3 y ; y ; 2)$ with $y \in P$.

It is interesting that in both cases $z$ is constant (and a prime number).