8. find the two smallest natural numbers $n$ so that the fractions

$$
\frac{68}{n+70}, \frac{69}{n+71}, \frac{70}{n+72}, \ldots, \frac{133}{n+135}
$$

are all irreducible.

## Solution
The fractions have the form $k /(k+n+2)$ for $68 \leq k \leq 133$. Because $\operatorname{ggT}(k, k+n+2)=$ $\operatorname{ggT}(k, n+2)$ they are all irreducible if $n+2$ is not divisible by $68, \ldots, 133$. So $n+2$ must not be divisible by any prime number $p$ that is a divisor of at least one of the numbers $68, \ldots, 133$. Because $133-68=65$, all prime numbers $p \leq 65$ are thus omitted as divisors of $n+2$, since they divide at least one of the numbers $68, \ldots, 133$. In addition, the prime numbers $68 \leq p \leq 133$ are of course also out of the question. The two smallest possibilities for $n+2$ are therefore $n+2=67$ and $n+2=137$, i.e. $n=65,135$.