Johan Gunardi wrote: Let $ f(x) = a_{2n}x^{2n} + a_{2n - 1}x^{2n - 1} + \cdots + a_1x + a_0$, with $ a_i = a_{2n - 1}$ for all $ i = 1,2,\ldots,n$ and $ a_{2n}\ne0$. Prove that there exists a polynomial $ g(x)$ of degree $ n$ such that $ g\left(x + \frac1x\right)x^n = f(x)$. The condition on $ a_i's$ must is $ a_i=a_{2n-i},i=\overline{0,2n}$.

With the new condition, it suffices to show $ x^n+x^{-n}$ is polynomial of $ x+x^{-1}$. This is trivial by induction with $ (x^n+x^{-n})=(x+x^{-1})(x^{n-1}+x^{1-n})-(x^{n-2}+x^{2-n})$.