Balkan Mathematical Olympiad

Consider the complex number $\omega =(a\pm i\sqrt{b}{)}^2-1$, where $a,b\in \mathbb{Z}$, $b\ge 0$. It is easy to see that $|\omega |=\sqrt{(a^2-b-1{)}^2+4a^{2}b}>1$ or $|\omega |=0$, where the latter is possible only when $a=0$ or $b=0$. Set in the condition $x=i\sqrt{2013}$ and consider $f_k(i\sqrt{2013})=a\pm i\sqrt{b}$. It follows from the above that one must have $f_k(i\sqrt{2013})=0$ for every $k=1,2,\dots ,n$. Then $(-1{)}^n+1=0$, i.e. $n$ must be odd. Let $n=2t+1$ be odd. We consider the sequence of polynomials $\{h_k(x){\}}_{k=1}^{\infty }$, defined by $h_1(x)=x^2+2013$, (we omit $x$ for a while) $h_{k+1}=4h_k^3-3h_k$ for $k\ge 1$. It is easy to see that $$h_{k+1}^2-1=(4h_k^3-3h_k{)}^2-1=(4h_k^2-1{)}^2(h_k^2-1).$$ We have $$\begin{array}{rl}{h}_{t+1}^2-1& =(4h_t^2-1{)}^2(h_t^2-1)=(4h_t^2-1{)}^2(4h_{t-1}^2-1{)}^2(h_{t-1}^2-1)\\ & =\cdots =(h_1^2-1)\prod _{m=1}^t(4h_m^2-1{)}^2\end{array}$$ and, on the other hand, $h_{t+1}^2(x)-1=(x^2+2013{)}^2{g}_{t+1}^2(x)$ since it obviously follows by induction that $h_k(x)=(x^2+2013)g_k(x)$ for every $k$ and some $g_k(x)\in \mathbb{Z}[x]$.