Bulgarian National Mathematical Olympiad

Let $\alpha_1 < \alpha_2 < \dots < \alpha_n$ be the roots of $f(x)$. Assume that
$$x(x+1)(x+2)(x+4)f(x) + a = g^k(x).$$
Note that $a = b^k = g^k(0)$.
If $k \ge 3$ is odd then the polynomial $g^k(x)-b^k$ has $n+4$ distinct real roots which will be also roots of $g(x)-b$. However, the degree of $g(x)-b$ is $(n+4)/k < n+4$, i.e. $g(x) = b$, which is impossible.
Now it is enough to prove that $k=2$ is also impossible. We have $a = b^2$, where we can assume that $b > 0$. Then
$$x(x+1)(x+2)(x+4)f(x) = g_1(x)g_2(x),$$
where $g_1(x) = g(x)+b$ and $g_2(x) = g(x)-b$. The roots of $g_1(x)$ and $g_2(x)$ are the numbers $-4, -2, -1, 0, \alpha_1, \dots, \alpha_n$. Since $g_1(x) > g_2(x)$ for every $x$, the number $-4$ is a root of $g_1(x)$. Since the derivatives of $g_1(x)$ and $g_2(x)$ coincide, the Rolle's theorem shows that $-2$ and $-1$ are roots of $g_2(x)$ while $0$ is a root of $g_1(x)$.
Let $g_1(x) = x(x+4) \prod_{j=1}^{s}(x - \alpha_j)$. Then
$$|g_1(-1)| = 3 \prod_{j=1}^{s}(1 + \alpha_j) < 4 \prod_{j=1}^{s}(2 + \alpha_j) = |g_1(-2)|,$$
which contradicts to $g_1(-1) = g_1(-2) = g(-1) + b = 2b$.