Balkan Mathematical Olympiad

The polynomial $P(x)$ can be written in the form $$P(x)=[(x-4{)}^2+3^2][(x-8{)}^2+6^2]\cdots [(x-4n{)}^2+9n^2]+1$$ Let we can write $P(x)$ in the form $P(x)=Q(x)R(x)$, where $$Q(x)=a_0+a_{1}x+\cdots +a_k{x}^k,1\le k<2n,a_0,a_1,\dots ,a_k\in \mathbb{Z},a_k\ne 0$$ $$R(x)=b_0+b_{1}x+\cdots +b_l{x}^l,1\le l<2n,b_0,b_1,\dots ,b_l\in \mathbb{Z},k+l=2n,b_l\ne 0.$$ We observe that for $x=x_{\pm m}\equiv 4m\pm 3mi=m(4\pm 3i)$, $m=1,2,\dots ,n$, $i^2=-1$, we have $$P(x_{\pm m})=1\iff Q(x_{\pm m})R(x_{\pm m})=1(1).$$ Since $Q(x),R(x)\in \mathbb{Z}[x]$ from (1) we have the following possible cases: $$Q(x_{\pm m})=\pm 1,R(x_{\pm m})=\pm 1\text{or}Q(x_{\pm m})=\pm i,R(x_{\pm m})=\mp i.$$ However, looking at the form of $Q(x_{\pm m})=Q(4m\pm 3mi)$ we observe that its imaginary part is a multiple of $3m$, which means that $Q(x_{\pm m})$ cannot obtain the values $\pm i$. Similarly, we observe the same for the polynomial $R(x)$. Therefore the possible cases are: $$Q(x_{\pm m})=1,R(x_{\pm m})=1\text{or}Q(x_{\pm m})=-1,R(x_{\pm m})=-1(2).$$ From relation (2) we conclude that the polynomial $$f(x):=Q(x)-R(x)$$ has the $2n$ roots $x_{\pm m}=4m\pm 3mi$, $m=1,2,\dots ,n$, while $\mathrm{deg}f(x)\le 2n-1$. Therefore the polynomial $f(x):=Q(x)-R(x)$ must be the zero polynomial and so $$Q(x)=R(x)=a_0+a_{1}x+\cdots +a_k{x}^k,k=n.$$ Finally, then we have $P(x)=(Q(x){)}^2=(a_0+a_{1}x+\cdots +a_n{x}^n{)}^2$, from which for $x=0$ we get $$\begin{array}{rl}P(0)=(Q(0){)}^2& \iff a_0^2=25\cdot 100\cdots 25n^2+1\\ & \iff a_0^2=5^{2n}(1\cdot 2\cdots n{)}^2+1\iff a_0^2=(5^n\cdot (n!){)}^2+1,\end{array}$$ which is absurd.