BULGARIAN NATIONAL MATHEMATICAL OLYMPIAD

Let ${\alpha }_1<{\alpha }_2<\cdots <{\alpha }_n$ 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 $g_1(-1)=g_1(-2)=g(-1)+b=2b$.