1. Determine all real numbers $c$ for which the equation

$$
\left(c^{2}+c-8\right)(x+2)-8|x-c+2|=c|x+c+14|
$$

has infinitely many solutions in the set of integers.
1. If we denote for a given real $c$

$$
f_{c}(x)=c|x+c+14|+8|x-c+2|-\left(c^{2}+c-8\right)(x+2)
$$

the corresponding piecewise linear function, it is clear that the equation $f_{c}(x)=0$ will have infinitely many integer solutions if and only if the function $f_{c}$ is identically zero on one of the infinite intervals $(-\infty, \min (c-2,-c-14)\rangle$ or $\langle\max (c-2,-c-14), \infty)$. We will investigate both possibilities step by step.

a) Let $x \leqq \min (c-2,-c-14)$, for such $x$ we have

$$
\begin{aligned}
f_{c}(x) & =-c(x+c+14)-8(x-c+2)-\left(c^{2}+c-8\right)(x+2)= \\
& =-c(2+c) x-3 c^{2}-8 c=-c(x(c+2)+3 c+8) .
\end{aligned}
$$

On this interval, the function $f_{c}$ will be identically zero if and only if $c=0$ (the system $c+2=0,3 c+8=0$ has no solution).

b) Let $x \geqq \max (c-2,-c-14)$, for such $x$ we have

$$
\begin{aligned}
f_{c}(x) & =c(x+c+14)+8(x-c+2)-\left(c^{2}+c-8\right)(x+2)= \\
& =\left(16-c^{2}\right) x-c^{2}+4 c+32 .
\end{aligned}
$$

On this interval, the function $f_{c}$ will be identically zero if and only if $c^{2}=16$ and $c^{2}-4 c-32=0$. Substituting $c^{2}=16$ into the second equation yields $c=-4$, which is clearly the only solution to both equations.

Conclusion. The given equation has infinitely many integer solutions if and only if $c=0$ or $c=-4$ (in the first case, the equation is satisfied by all integers $x \leqq-14$, and in the second case by all integers $x \geqq-6$).

For a complete solution, 6 points are awarded, 2 points for the idea that the given function must be zero on an unbounded interval, 2 points for each of the two possibilities.