多项式$f(x)$整除$f(x^2 + x + 1)$

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(Math Magazine Q718, Bjorn Poonen) Suppose $f(x)$ and $g(x)$ are nonzero real polynomials satisfying
\[f(x^2 + x + 1) = g(x)f(x).\]
Show that $f(x)$ has even degree.
来自FB

例如$f(x)=x^2 + 1$. $g(x)=x^2 + 2x + 2.$
$f(x)=x^2 + 2x + 3$. $g(x)=x^2 + 2.$

@Andrea Aurigemma的证明
Assume $x_0$ is a real root of $f(x)$. Then by the functional equation, $x_0^2+x_0+1$ is also a root of $f(x)$.
But the real function $h(x)=x^2+x+1$ satisfies $h(x)>x$. Therefore from a real root $x_0$ we could generate infinite distinct roots $h(x_0),h(h(x_0)),\cdots$ , but this is impossible. So $f(x)$ has even degree because every odd degree real polynomial has at least a real root.

假设 $x_0$ 是 $f(x)$ 的实根。由函数方程，$x_0^2+x_0+1$也是$f(x)$的根。
函数$h(x)=x^2+x+1$满足$h(x)>x$。因此，从实根 $x_0$ 我们可以生成无限不同的根 $h(x_0),h(h(x_0)),\cdots$ ，这是不可能的。
所以 $f(x)$ 具有偶数次，因为每个奇数次实数多项式至少有一个实根。