Find all real coefficient polynomial with the degree is $3$,and such $$f(x)|f(x^2-2),\forall x\in Z $$
I try: let $f(x)=ax^3+bx^2+cx+d$,then $$f(x^2-2)=ax^6-6ax^4+12ax^2-8a+bx^4-4bx^2+4b+cx^2-2c+d$$,then such $$(ax^3+bx^2+cx+d)|ax^6-6ax^4+12ax^2-8a+bx^4-4bx^2+4b+cx^2-2c+d,\forall x\in Z $$following I can't it
Suppose $f(x) \mid f(x^2 - 2)$. Ghartal's argument has already shown that if $x$ is a root of $f$, then $x^2 - 2$ is also a root of $f$. In fact, by the same argument, we have that for all $x \in \mathbb{C}$, the multiplicity of $x^2 - 2$ as a root of $f$ is at least the multiplicity of $x$ as a root of $f$.
I claim that the above condition is also sufficient to have $f(x)\mid f(x^2 - 2)$. Indeed, we may write $f$ as $$ f(x) = k\prod_{c \in \mathbb{C}} (x - c)^{m_c} $$ for some $k \in \mathbb{C}$ (w.l.o.g. $k \neq 0$) and where $m_c \in \mathbb{N}$ is non-zero for finitely many $c \in \mathbb{C}$. Then $$ f(x^2 - 2) = k\prod_{c \in \mathbb{C}} (x^2 - 2 - c)^{m_c} . $$ Now by hypothesis we have $m_{c^2 - 2} \geq m_{c}$ for all $c \in \mathbb{C}$. Hence $(x -c)^{m_c} \mid (x^2 - 2 - (c^2 - 2))^{m_{c^2 - 2}}$ for all $c \in \mathbb{C}$. In particular, $f(x) \mid f(x^2 - 2)$. This proves the claim.
To find all degree three polynomials (in $\mathbb{R}[X]$, or $\mathbb{C}[X]$ if you wish) with the given property, it suffices to "trace back" the function $x \mapsto x^2 - 2$ three steps and then take products of the induced linear polynomials; respecting the claim above. The function $x \mapsto x^2 - 2$ does: $$ +/- \sqrt{3} \mapsto 1 \mapsto - 1 \qquad \text{and} \qquad 0 \mapsto -2 \mapsto 2. $$ so the possible polynomials of degree at most three are (up to multiplying by a constant):
The polynomials of degree exact three are the last 11.
Hint: Write $$f(x^2-2)=g(x)f(x)$$ If $x$ is a root of $f$ then $x^2-2$ is a root too. Then $(x^2-2)^2-2$ is a root of $f$ and etc. But $f$ cannot have infinitely many roots.
