4. Find all rational numbers $r$ and all integers $k$ that satisfy the equation

$$
r(5 k-7 r)=3
$$
$\mathbf{I} / 4$. Clearly, $r \neq 0$. Let's write $r$ in the form of a reduced fraction $r=\frac{m}{n}$ and assume that $n$ is a natural number. Then we get $\frac{m}{n}\left(5 k-7 \frac{m}{n}\right)=3$ or $m(5 k n-7 m)=3 n^{2}$. Therefore, $m$ is a divisor of $3 n^{2}$. Since $m$ and $n$ are coprime, $m \mid 3$. We consider four cases.

If $m=1$, we get $5 k n-7=3 n^{2}$ or $n(5 k-3 n)=7$, from which it follows that $n \mid 7$. Since $n$ is a natural number, it is either 1 or 7. For $n=1$, we get the solution $k=2$ and $r=1$, and for $n=7$, we get the contradictory equation $5 k=22$.

If $m=3$, we get $n(5 k-n)=21$. Again, we see that $n \mid 21$, but since $m$ and $n$ are coprime, $n \mid 7$. Again, $n=1$ or $n=7$. We get a solution only in the second case, namely $k=2$ and $r=\frac{3}{7}$.

If $m=-1$, we get $n(5 k+3 n)=-7$, so again $n \mid 7$. For $n=1$, we get the solution $k=-2$ and $r=-1$, and for $n=7$, there is no solution.

Finally, if $m=-3$, we get $n(5 k+n)=-21$. Since $m$ and $n$ are coprime, $n$ is a divisor of 7. We get one more solution, namely $k=-2$ and $r=-\frac{3}{7}$.

All pairs of solutions $(k, r)$ are $(2,1),(-2,-1),\left(2, \frac{3}{7}\right)$ and $\left(-2,-\frac{3}{7}\right)$.

![](https://cdn.mathpix.com/cropped/2024_06_07_2dd3bccc19bc96d6d7dfg-08.jpg?height=54&width=1636&top_left_y=790&top_left_x=216)

![](https://cdn.mathpix.com/cropped/2024_06_07_2dd3bccc19bc96d6d7dfg-08.jpg?height=51&width=1633&top_left_y=840&top_left_x=220)
Limiting the possibilities for the number $n$ in all four cases (i.e., $n \mid 7$ or $n \mid 21$) ... 1 point Each pair of solutions $(k, r) \in\left\{(2,1),(-2,-1),\left(2, \frac{3}{7}\right),\left(-2,-\frac{3}{7}\right)\right\} \ldots \ldots \ldots \ldots \ldots \ldots$ 1 point each (If only the solutions are given and there is no process showing that there are no other solutions, award a maximum of 4 points.)