Czech-Polish-Slovak Match

We check that for $n=3$ $$x^3-3x^2+2x+6=(x+1)(x^2-4x+6).$$ Suppose that for $n=4$ we have $$x^4-3x^3+2x^2+6=(x^2+ax+b)(x^2+cx+d).$$ Then, comparing the coefficients we obtain $$a+c=-3,ac+b+d=2,bd=6.$$ The first equation implies that $a$ and $c$ are of different parity and therefore, by the second equality, $b$ and $d$ are of the same parity. This contradicts the third equation. Hence we may assume that $n\ge 5$. Suppose that we have $$W(x)=P(x)Q(x),(1)$$ where $$P(x)=a_k{x}^k+a_{k-1}{x}^{k-1}+\cdots +a_{1}x+a_0,$$ $$Q(x)=b_{n-k}{x}^{n-k}+b_{n-k-1}{x}^{n-k-1}+\cdots +b_{1}x+b_0,$$ and $a_k=b_{n-k}=\pm 1$. With no loss of generality we may assume that $k\le \lfloor \frac{1}{2}n\rfloor <n-2$ (because $n\ge 5$). Comparing the coefficients of both sides of (1) we obtain the following system of equations: $$\begin{array}{rl}{a}_0{b}_0& =6,\\ a_0{b}_1+a_1{b}_0& =0,\end{array}$$ $$a_0{b}_k+a_1{b}_{k-1}+\cdots +a_{k-1}{b}_1+a_k{b}_0=0.$$ Now we can easily prove by induction that $a_0$ divides $a_1,a_2,\dots ,a_k$. Indeed, having established this fact for $a_1,a_2,\dots ,a_l$ we write $$\begin{array}{rl}0& =a_0(a_0{b}_{l+1}+a_1{b}_l+\cdots +a_l{b}_1+a_{l+1}{b}_0)=\\ & =a_0^2{b}_{l+1}+a_0{a}_1{b}_l+\cdots +a_0{a}_l{b}_1+6a_{l+1},\end{array}$$ and hence $$6a_{l+1}=-(a_0^2{b}_{l+1}+a_0{a}_1{b}_l+\cdots +a_0{a}_l{b}_1).$$ We note that all the summands on the right hand side are divisible by $a_0^2$, therefore the left hand side also has this property. Hence $a_0\mid a_{l+1}$. But we have $a_k=\pm 1$; hence $a_0=\pm 1$ and with no loss of generality we may take $a_0=1$ and then we have $b_0=6$. Now if we repeat the above arguments for the coefficients of $Q$, then we see that $b_0=6$ divides $b_1,b_2,\dots ,b_{n-3}$ (we set, if necessary, $b_l=0$ if $l>n-k$). Then we obtain a contradiction ($b_{n-k}=\pm 1$), unless $n-k>n-3$. We consider two cases. The case of $k=2$. Then we arrive at $$a_0{b}_{n-k}+a_1{b}_{n-k-1}+\cdots +a_{n-k-1}{b}_1+a_{n-k}{b}_0=2,$$ a contradiction, because on the left hand side all the summands, except for the first one, are divisible by 6 and hence even, and the first summand is equal to $\pm 1$. The case of $k=1$. Then the problem reduces to finding an integer root of the polynomial $W$; it is easy to check that if $n$ is even there are no such roots; if $n$ is odd, $W(-1)=0$. Therefore the answer to the problem is: $n$ is odd.