VMO

Clearly, polynomial $P(x)$ has degree $4n$ and has no real root. Thus, any factor of $P(x)$ has even degree. Suppose that $P(x)$ can be expressed as the product of $n+1$ polynomials with degree greater than 0, so $$P(x)=P_1(x)\cdot P_2(x)\cdots P_{n+1}(x)$$ then $P_i(x)$ with $i=1,2,\dots ,n$ has even degree. Since the sum of degrees of $P_i(x)$ is $4n$ then there are at least two polynomials, let say $P_1(x),P_2(x)$ of degree 2. Since $P(x)$ has the leading coefficient 1, suppose that $P_1(x),P_2(x)$ have leading coefficients 1, which are $$P_1(x)=x^2+ax+b,P_2(x)=x^2+cx+d\text{ where }a,b,c,d\in \mathbb{Z}.$$ Since $P_1(x),P_2(x)$ have no real roots, we have $P_1(x)>0,P_2(x)>0$ for all integers $x$. We have $$13=P(1)=P_1(1)\cdot P_2(1)\cdots P_{n+1}(1)$$ and $$13=P(6)=P_1(6)\cdot P_2(6)\cdots P_{n+1}(6).$$ From this, at least one of two numbers $P_1(1)$ and $P_2(1)$ equal 1. Without loss of generality, we assume that $P_1(1)=1$. It follows that $a=-b$, so $P_1(6)=36-5b$. We observe that $36-5b>0$ and cannot be equal to 13 so $36-5b=1$, thus $b=7,a=-7$. But in this case the polynomial $P_1(x)=x^2-7x+7$ has real root, which is a contradiction. Therefore, the above assumption is wrong so $P(x)$ cannot be expressed as the product of $(n+1)$ non-constant polynomials. $\square$