59th Ukrainian National Mathematical Olympiad

First, we go through cases when this equation is not quadratic. For $k=0$, equation becomes $-4x+6=0$, and has a non-integer solution. For $k=2$, equation becomes $-6x+6=0$, and has a single solution $x=1$, which satisfies the statement.

Now, suppose $k(2-k)\ne 0$. We find the discriminant of this quadratic equation: $$D=(k^2+8k+16)-24(2k-k^2)=25k^2-40k+16=(5k-4{)}^2$$ So the roots are: $$x_1,x_2=\frac{(k+4)\pm (5k-4)}{2k(2-k)}$$ Calculating both roots: $$x_1=\frac{(k+4)+(5k-4)}{2k(2-k)}=\frac{6k}{2k(2-k)}=\frac{3}{2-k}$$ $$x_2=\frac{(k+4)-(5k-4)}{2k(2-k)}=\frac{-4k+8}{2k(2-k)}=\frac{2}{k}$$ We must find all cases when this is a positive integer. Let $\frac{2}{k}=m$ be a positive integer, then $k=\frac{2}{m}$.

Then: $$\frac{3}{2-k}=\frac{3}{2-\frac{2}{m}}=\frac{3m}{2(m-1)}$$ We require $\frac{3m}{2(m-1)}$ to be a positive integer. Since $m$ is a positive integer, $m$ and $m-1$ are coprime. Hence, the last condition is only possible for $m-1=1\Rightarrow k=1$ or $m-1=3\Rightarrow k=\frac{1}{2}$.

Therefore, the real values of $k$ are $k=2$, $k=1$, and $k=\frac{1}{2}$.