## Task 12/69

All pairs $(x ; y)$ are to be found that satisfy the system of equations

$$
\begin{aligned}
& |y-x|=|x+1| \\
& \frac{y-3}{4}=\left[\frac{x-1}{5}\right]
\end{aligned}
$$

where $[a]$ is an integer with $a-1<[a] \leq a$.
We eliminate the absolute value signs in (1) by case distinction:

1st Case: $\operatorname{sgn}(y-x)=\operatorname{sgn}(x+1)$. Then $y-x=x+1, y=2 x+1$

2nd Case: $\operatorname{sgn}(y-x) \neq \operatorname{sgn}(x+1)$. Then $y-x=-x-1, y=-1$

We rearrange (2) and substitute: $y=4\left[\frac{x-1}{5}\right]+3$.

1st Case:

$$
2 x+1=4\left[\frac{x-1}{5}\right]+3 \rightarrow \frac{x-1}{2}=\left[\frac{x-1}{5}\right]
$$

This implies that $x$ is odd, $x=2 k+1$ ( $k$ integer). Thus we get

$$
\frac{x-1}{5}-1<\frac{x-1}{2} \leq \frac{x-1}{5} \rightarrow 2 x-12<5 x-5 \leq 2 x-2 \rightarrow-12<3(2 k+1)-5 \leq-2 \rightarrow
$$

$$
\rightarrow-10<6 k \leq 0 \rightarrow-\frac{5}{3}<k \leq 0
$$

thus (due to the integrality of $k$) $k_{1}=-1, k_{2}=0$. Since all transformations are reversible, this yields the solution pairs $(-1 ;-1)$ and $(1 ; 3)$.

2nd Case:

$$
-1=4\left[\frac{x-1}{5}\right]+3 \rightarrow \frac{x-1}{5}-1<-1 \leq \frac{x-1}{5} \rightarrow-4 \leq x<1
$$

Since all transformations are reversible, we obtain the solutions $\left(x_{0} ;-1\right)$ with $-4 \leq x_{0}<1$, which includes the already determined pair $(-1 ;-1)$.

The sought pairs are thus $(1 ; 3)$ and $\left(x_{0} ;-1\right)$ with $-4 \leq x_{0}<1$.