58. National mathematical olympiad Final round

Let $\mathrm{deg}(f_k)={\alpha }_k\in \mathbb{N},1\le k\le n$. We consider all indexes modulo $n$. The condition implies the equalities ${\alpha }_k+{\alpha }_{k+1}={\alpha }_{k+1}{\alpha }_{k+2}$, i.e. ${\alpha }_{k+1}$ divides ${\alpha }_k$ for every $k=1,2,\dots ,n$. Therefore ${\alpha }_1={\alpha }_2=\cdots ={\alpha }_n=2$.

Let $f(x)=a_k{x}^2+b_{k}x+c_k,1\le k\le n$, where $a_k\ne 0$. Comparing the coefficients of $x^4$ we obtain $a_k=a_{k+2}^2,1\le k\le n$. If $n=2m$ is even, then $a_1=a_3^2=a_5^2=\cdots =a_{2m-1}^2=a_1^{2m}$ and therefore $a_1=a_3=\cdots =a_{2m-1}=1$. We analogously have $a_2=a_4=\cdots =a_{2m}=1$. If $n$ is odd, the same argument shows that $a_1=a_2=\cdots =a_n=1$.

Comparing the coefficients of $x^3$ we obtain the equalities $b_k+b_{k+1}=2b_{k+2}$, $1\le k\le n$. Let $\min \{b_1,b_2,\dots ,b_n\}=b_s:=b$. Then $b_{s-2}+b_{s-1}=2b_s$ shows that $b_{s-2}=b_{s-1}=b$ and, continuing in the same way, we conclude that $b_1=b_2=\cdots =b_n=b$. We now compare the coefficients of $x^2$ and get $c_k+c_{k+1}=2c_{k+2}+b$, $1\le k\le n$. Summing up these equalities we obtain $nb=0$, i.e. $b=0$. Then the same argument as above gives $c_1=c_2=\cdots =c_n=c$. Hence $f(x)=x^2+c$ and the equations are $(x^2+c)(x^2+c)=(x^2+c{)}^2+c$. Then $c=0$ and $f_1(x)=f_2(x)=\cdots =f_n(x)=x^2$.