Slovenia — National Math Olympiad

Obviously, $r \ne 0$. Let us write $r$ as a reduced fraction $r = \frac{m}{n}$ and let us assume that $n$ is a positive integer. Then $\frac{m}{n}(5k - 7m) = 3$ or, equivalently, $m(5kn - 7m) = 3n^2$. Hence, $m$ divides $3n^2$. Since $m$ and $n$ are coprime, we conclude that $m$ divides $3$. Let us consider four cases.

If $m = 1$ we have $5kn - 7 = 3n^2$ or $n(5k - 3n) = 7$, which implies that $n$ divides $7$. Since $n$ is a positive integer, it equals either $1$ or $7$. When $n=1$ we get $k=2$ and $r=1$. When $n=7$ the equation $5k=22$ gives us no integer solutions.

If $m=3$ we have $n(5k-n) = 21$. We see that $n$ divides $21$. Again, $m$ and $n$ are coprime, so $n$ divides $7$. Once more we have either $n=1$ or $n=7$. This time we obtain the solution only in the second case: $k=2$, $r=\frac{3}{7}$.

If $m=-1$ we have $n(5k+3n) = -7$, so $n$ divides $7$. When $n=1$ the solution is $k=-2$, $r=-1$. When $n=7$ there are no solutions.

If $m=-3$ we have $n(5k+n) = -21$. As $m$ and $n$ are coprime, $n$ divides $7$. We find one last solution, $k=-2$, $r=-\frac{3}{7}$.

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