SAPOSTO wrote: Let $ n\ge 3$ be a natural number. Find all nonconstant polynomials with real coeficcietns $ f_{1}\left(x\right),f_{2}\left(x\right),\ldots,f_{n}\left(x\right)$, for which $ f_{k}\left(x\right)f_{k + 1}\left(x\right) = f_{k + 1}\left(f_{k + 2}\left(x\right)\right)$,$ 1\le k\le n$, for every real $ x$ (with $ f_{n + 1}\left(x\right)\equiv f_{1}\left(x\right)$ and $ f_{n + 2}\left(x\right)\equiv f_{2}\left(x\right)$). Let $ d_k>0$ the degree of $ f_k(x)$. The equation implies $ d_k+d_{k+1}=d_{k+1}d_{k+2}$ and so $ d_k=d_{k+1}(d_{k+2}-1)$ If $ d_{k+2}=1$, $ d_k=0$, which is impossible. So $ d_{k+2}>1$ and so $ d_k\geq d_{k+1}$ So, the cyclic equality implies $ d_k=$constant, and so $ d_k=2$ $ \forall k$ So $ f_k(x)=a_kx^2+b_kx+c_k$ and the equation becomes : $ (a_kx^2+b_kx+c_k)(a_{k+1}x^2+b_{k+1}x+c_{k+1})$ $ =a_{k+1}(a_{k+2}x^2+b_{k+2}x+c_{k+2})^2+$ $ b_{k+1}(a_{k+2}x^2+b_{k+2}x+c_{k+2})$ $ +c_{k+1}$ Equating coefficients of $ x^4$ in this equality implies $ a_ka_{k+1}=a_{k+1}a_{k+2}^2$ and so $ a_k=a_{k+2}^2$ (since $ a_{k+1}\neq 0$, else $ d_k=0$). And so $ a_k=a_k^{2^n}$ and so $ a_k=1$ and the equation becomes : $ (x^2+b_kx+c_k)(x^2+b_{k+1}x+c_{k+1})$ $ =(x^2+b_{k+2}x+c_{k+2})^2+$ $ b_{k+1}(x^2+b_{k+2}x+c_{k+2})$ $ +c_{k+1}$ Equating coefficients of $ x^3$ in this equality implies $ b_k+b_{k+1}=2b_{k+2}$ and so $ b_k=b_{k+1}$ else $ b_{n+1}\neq b_1$. So $ b_k=b$ $ \forall k$ and the equation becomes : $ (x^2+bx+c_k)(x^2+bx+c_{k+1})$ $ =(x^2+bx+c_{k+2})^2+$ $ b(x^2+bx+c_{k+2})$ $ +c_{k+1}$ Equating coefficients of $ x^2$ in this equality implies $ c_k+c_{k+1}+b^2=b^2+2c_{k+2}+b$ and so $ c_k+c_{k+1}=2c_{k+2}+b$ Adding the $ n$ equalities $ c_k+c_{k+1}=2c_{k+2}+b$ implies $ b=0$ and then $ c_k=c$ constant and the equation becomes : $ (x^2+c)(x^2+c)$ $ =(x^2+c)^2+c$ and so $ c=0$ So, the only solution is $ f_k(x)=x^2$