Mathematical Reflections 2015, Issue 3 - Problem U337

Problem:
Let $n$ be a positive integer. Find all real polynomials $f$ and $g$ such that $$(x^2+x+1)^n f(x^2-x+1)=(x^2-x+1)^n g(x^2+x+1),$$ for all real numbers $x$.

Proposed by Marcel Chirita, Bucharest, Romania 

Solution:
Observe that $x^2-x+1=(1-x)^2-(1-x)+1$. Hence, by substitution $x \mapsto 1-x$, we get $$(x^2-3x+3)^n f(x^2-x+1)=(x^2-x+1)^n g(x^2-3x+3)$$ and dividing by the original equation, we have
\begin{equation}\label{first-eq}
\dfrac{g(x^2-3x+3)}{g(x^2+x+1)}=\left(\dfrac{x^2-3x+3}{x^2+x+1}\right)^n.             (1)
\end{equation}
Moreover, since $x^2+x+1=(-1-x)^2-(-1-x)+1$, by substitution $x \mapsto -1-x$, we get $$(x^2+x+1)^n f(x^2+3x+3)=(x^2+3x+3)^n g(x^2+x+1)$$ and dividing by the original equation, we have $$\dfrac{f(x^2+3x+3)}{f(x^2-x+1)}=\left(\dfrac{x^2+3x+3}{x^2-x+1}\right)^n.$$
By substitution $x \mapsto -x$, we obtain
\begin{equation}\label{second-eq}
\dfrac{f(x^2-3x+3)}{f(x^2+x+1)}=\left(\dfrac{x^2-3x+3}{x^2+x+1}\right)^n.            (2)
\end{equation}
Comparing (1) and (2), we obtain $$\dfrac{f(x^2-3x+3)}{g(x^2-3x+3)}=\dfrac{f(x^2+x+1)}{g(x^2+x+1)}$$ for all $x \in \mathbb{R}$. It follows that $\dfrac{f(x^2-3x+3)}{g(x^2-3x+3)}$ and $\dfrac{f(x^2+x+1)}{g(x^2+x+1)}$ have the same derivative for all $x \in \mathbb{R}$. So,
$$\begin{array}{lll} & & (2x-3)\dfrac{f'(x^2-3x+3)g(x^2-3x+3)-f(x^2-3x+3)g'(x^2-3x+3)}{(g(x^2-3x+3))^2}=\\&=&(2x+1)\dfrac{f'(x^2+x+1)g(x^2+x+1)-f(x^2+x+1)g'(x^2+x+1)}{(g(x^2+x+1))^2}.\end{array}$$ for all $x \in \mathbb{R}$. Since $(1-x)^2+(1-x)+1=x^2-3x+3$, by substition $x \mapsto 1-x$ into the right-hand side, we have $$(2x-3)\dfrac{f'(x^2-3x+3)g(x^2-3x+3)-f(x^2-3x+3)g'(x^2-3x+3)}{g^2(x^2-3x+3)}=0$$ for all $x \in \mathbb{R}$. It follows that $\dfrac{f(x^2-3x+3)}{g(x^2-3x+3)}$ is constant on $\mathbb{R}$, i.e. $f(x)=\lambda g(x)$ for some $\lambda \in \mathbb{R}$. Substituting into the original equation, we get $\lambda=1$.
Now, from equation (2), we get
$$\dfrac{f(x^2-3x+3)}{(x^2-3x+3)^n}=\dfrac{f(x^2+x+1)}{(x^2+x+1)^n}$$ for all $x \in \mathbb{R}$. It follows that $\dfrac{f(x^2-3x+3)}{(x^2-3x+3)^n}$ and $\dfrac{f(x^2+x+1)}{(x^2+x+1)^n}$ have the same derivative for all $x \in \mathbb{R}$. So,
$$\begin{array}{lll} & & (2x-3)\dfrac{f'(x^2-3x+3)(x^2-3x+3)-nf(x^2-3x+3)}{(x^2-3x+3)^{n+1}}=\\&=&(2x+1)\dfrac{f'(x^2+x+1)(x^2+x+1)-nf(x^2+x+1)}{(x^2+x+1)^{n+1}}.\end{array}$$ for all $x \in \mathbb{R}$. By substitution $x \mapsto 1-x$ into the right-hand side, we get once again that $$\dfrac{d}{dx} \dfrac{f(x^2-3x+3)}{(x^2-3x+3)^n}=0$$ for all $x \in \mathbb{R}$, so $f(x)=cx^n$ for some $c \in \mathbb{R}$. It follows that $f(x)=g(x)=cx^n$, where $c \in \mathbb{R}$.