4. The condition means that the polynomial $x^{4}+5 x^{3}+a x^{2}+b x+c$ is divisible by the polynomial of the third degree $x^{3}+(8+b) x^{2}+(b+4) x+(c+3)$ without a remainder. This condition is equivalent to the system

$$
\left\{\begin{array}{l}
a-b-4=(8+b)(-3-b) \\
b-(c+3)=(4+b)(-3-b) \\
c=(c+3)(-3-b)
\end{array}\right.
$$

From the second equation, we express $c+3$ and substitute it into the right-hand side of the third: $c=\left(b^{2}+8 b+12\right)(-3-b)$. Now we add this equation to the second and obtain an equation for $b: b^{3}+12 b^{2}+44 b+45=0$. From this, the only integer solution is $b=-5$. Consequently, $a=5$, $c=-6$. It is easy to verify that the equation $x^{3}+3 x^{2}-x-3=0$ has three solutions $\pm 1,-3$.
Answer: $a=5, b=-5, c=-6$.

Answer to variant: 3-2: $a=5, b=-6, c=-5$.