## Task 2 - 110922

Determine all ordered pairs $(a, b)$ of integers $a$ and $b (b \neq 0)$ with the following property: If the numerator $a$ of the fraction $\frac{a}{b}$ is replaced by the sum of $a$ and a suitable natural number $n (n \neq 0)$, and at the same time the denominator $b$ of this fraction is replaced by the product of $b$ and the same number $n$, the resulting fraction is equal to the fraction $\frac{a}{b}$ mentioned at the beginning.
Sought are number triples $(a, b, n)$, which satisfy the equation

$$
\frac{a}{b}=\frac{a+n}{b n}
$$

$b$ can be canceled and is thus arbitrarily selectable. After rearrangement, we obtain $(a-1)(n-$ $1)=1$.

Since solutions in $\mathbb{Z}$ are sought, it follows that $a-1=n-1= \pm 1$. Therefore, due to $n \neq 0$, the triples are given by $(2, b, 2), b \in \mathbb{Z} \backslash\{0\}$.

Second solution:

Assume that $(a, b)$ is a number pair that, together with a suitable number $n$, satisfies the given condition, then

$$
\frac{a+n}{b n}=\frac{a}{b}
$$

From this, we obtain $a b+b n=a b n$. Since $b \neq 0$, it follows that $a+n=a n$, thus

$$
a=n(a-1)
$$

From (1) it follows that $a \neq 1$ and therefore

$$
n=\frac{a}{a-1}=1+\frac{1}{a-1}
$$

Since $n$ is an integer, it further follows that

$$
a-1= \pm 1, \quad \text { so } a=2 \quad \text { or } \quad a=0
$$

and because $n>0$, finally $a=2$. Therefore, only number pairs of the form $(2, b)$ and for each of these pairs only $n=2$ can satisfy the conditions of the problem.

Indeed, $\frac{2+2}{b \cdot 2}=\frac{4}{2 b}=\frac{2}{b}$.

Thus, the solution set consists of all number pairs of the form $(2, b)(b \neq 0$, integer $)$.

Adapted from $[2]$