6. determine all odd natural numbers of the form

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

where $p>q$ are prime numbers.

## 1st solution
Let $n$ be this number. According to the prerequisite, $n(p-q)=p+q=(p-q)+2 q$, therefore $p-q$ is a divisor of $2 q$. Because $\operator name{ggT}(p-q, q)=1$, $p-q=1$ or $p-q=2$ therefore applies. In the first case, $p$ or $q$ is even, i.e. $p=3, q=2$ and $n=5$. In the second case, $p$ and $q$ are odd and therefore $n=\frac{p+q}{p-q}=\frac{2 q+2}{2}=q+1$ is even, contradiction. The only such number is therefore $n=5$.

## 2nd solution

Again, we consider the equation $n(p-q)=p+q$. We first assume that $p$ and $q$ are odd. $\operatorname{In}$ case $p \equiv q(\bmod 4)$ the left side of the equation is divisible by 4, but the right side is not. $\operatorname{In}$ case $p \not \equiv q(\bmod 4)$ the right side is divisible by 4, but the left side is not because $n$ is odd. This contradiction shows $q=2$. In this case, $n=\frac{p+2}{p-2}=1+\frac{4}{p-2}$, so $p-2$ is a divisor of 4 and therefore $p=3$ and $n=5$.

## 3rd solution

Transforming the equation from the second solution gives $p(n-1)=q(n+1)$. Since $p$ and $q$ are divisors, $n+1=a p$ and $n-1=b q$ for natural numbers $a, b$. Substitution shows $a=b$ and therefore $a(p-q)=(n+1)-(n-1)=2$. Again, $p-q=1$ or $=2$ and you can conclude as in the first solution.