## Task $5 / 75$

We are looking for all triples $(x ; y ; z)$ of prime numbers for which the equation $\frac{x+y}{x-y}=z$ holds.
From the given equation, it follows:

1) $x>y$ (from $x>y$ it follows that $z<0$, from $x=y$ it follows that $z$ does not exist)
2) $z \neq 2$ (from $z=2$ it follows that $x+y=2x-2y$, thus $x=3y$, which contradicts the requirement that $\mathrm{x}$ and $\mathrm{y}$ are prime numbers).

Since $z \neq 2$, $z=2k+1$ with $k \in \mathbb{N}$. This leads to

$$
x+y=(2k+1)(x-y)=2kx+x-2ky-y \rightarrow 2kx=2(k+1)xy \rightarrow x=y+\frac{y}{k}
$$

Since $x$ is a prime number and must be an integer, and $y$ is also a prime number, the only possible values for $k$ are initially $k=1$ and $k=y$. The value $k=1$ is excluded, as it would result in $x=2y$, which contradicts the primality of $\mathrm{x}$ and $\mathrm{y}$.

For $k=y$, it follows that $x=y+1$. The only consecutive prime numbers are 2 and 3, so $y=2$ and $x=3$ must be true. For these two values, $z$ is also a prime number:

$$
z=\frac{x+y}{x-y}=5
$$

The triplet $(x; y; z)=(3; 2; 5)$ is thus the only one that satisfies the given conditions.