## Task 4 - 201234

Determine all integers $k$ for which the equation

$$
\frac{x}{k-4}+\frac{k}{2(k-4)}+\frac{k+4}{x}=0
$$

is solvable (i.e., has at least one solution $x$), where all solutions $x$ are integers.
It is clear that we are dealing with the zeros of

$$
f_{k}(x)=\frac{2 x^{2}+k x+2(k-4)(k+4)}{2(k-4) x}, \quad k \in \mathbb{Z} \backslash\{4\}, x \neq 0
$$

We set the numerator to zero:

$$
\begin{aligned}
2 x^{2}+k x+2(k-4)(k+4) & =0 \\
x^{2}+\frac{k}{2} x+k^{2}-16 & =0 \\
x & =-\frac{k}{4} \pm \sqrt{\frac{k^{2}}{16}-k^{2}+16} \\
x & =-\frac{k}{4} \pm \frac{1}{4} \sqrt{-15 k^{2}+256}
\end{aligned}
$$

Thus, we obtain solutions that are integers for $k \in\{-4 ;-3 ;-2 ; 0 ; 2 ; 3\}$.