How can I prove that $g(\zeta)\in\mathbb R\implies g(\zeta)=h(\zeta+\bar\zeta)$ How can I prove that if $g(X)\in \mathbb Q[X]$ and $\zeta\in\mathbb C\backslash \mathbb R$, therefore $$g(\zeta)\in\mathbb R\implies g(\zeta)=h(\zeta+\bar\zeta)$$ for a certain polynomial $h(X)\in\mathbb Q[X]$ ?
Let $$g(X)=a_0+a_1X+...+a_nX^n,$$
I know that $$g(\zeta)\in\mathbb R\implies g(\zeta)=\overline{g(\zeta)}\underset{a_i\in\mathbb Q}{=}g(\bar\zeta)$$
therefore $$a_1(\zeta-\bar\zeta)+a_2(\zeta^2-\bar\zeta^2)+...+a_n(\zeta^n-\bar\zeta^n)=0,$$
and thus $$\Im(\zeta^k-\bar\zeta^k)=0,$$
for all $k$, but I don't know how to continue.
 A: The claim seems false as currently stated. Consider for example 
$g(X)=X^2, \zeta=i\sqrt[4]{7}$. Then $g(\zeta)=-\sqrt{7}\in{\mathbb R}$,
but $g(\zeta)$ can never equal $h(\zeta+\bar{\zeta})$ for any 
$h\in{\mathbb Q}[X]$ (because $g(\zeta)$ is irrational and
$h(\zeta+\bar{\zeta})$ is rational, indeed $\zeta+\bar{\zeta}=0$).
A: In the original contest question it was assumed that $|\zeta|=1$.
The question was asked in the KöMaL Magazine. You can find the solution at
http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=A498&l=en
A: I present my ideas for the case $\zeta\bar\zeta\in\Bbb Q$, which is more general than $\zeta\bar\zeta=|\zeta|^2=1$.  
It seems the following.
Since $g(\zeta)=g(\bar\zeta)$, $g(\zeta)+g(\bar\zeta)=2g(\zeta)\in\Bbb R$. So $$2a_0+a_1(\zeta+\bar\zeta)+a_2(\zeta^2+\bar\zeta^2)+...+a_n(\zeta^n+\bar\zeta^n)\in \Bbb R.$$
Now it rests to show that $\zeta^k+\bar\zeta^k$ is a polynomial with rational coefficiemts of $\zeta+\bar\zeta$ for each $k$, that can be done inductively, because $\zeta^0+\bar\zeta^0=2$, $\zeta^1+\bar\zeta^1=\zeta+\bar\zeta$ and  $$\zeta^{k+1}+\bar\zeta^{k+1}=(\zeta^{k}+\bar\zeta^{k})(\zeta+\bar\zeta)-\zeta\bar\zeta(\zeta^{k-1}+\bar\zeta^{k-1})$$ for each $k\ge 1$.
