Find all $f \in \Bbb Z[x]$ such that $x^6+x^3f''(x)=f(x^3)+x^3$ holds

Question

Find all $f \in \Bbb Z[x]$ such that $x^6+x^3f''(x)=f(x^3)+x^3$ holds.

My try:

I tried to find the degree of polynomial. Let $n$ be the degree of $f(x)$. $n$ has to be at least $3$ because $f(x)$ has second derivation.

• For $x^6+x^3f''(x)$, the degree $\le$ max{$6,3+n-2$}=max{$6,n+1$}

• For $f(x^3)+x^3$, the degree $\le$ max{$3n,3$}

I can't really learn much from this so I will just let this be.

Now, I tried to find the roots:

For $x=0 \rightarrow 0^6+0^3f''(0)=f(0^3)+0^3 \rightarrow f(0)=0$. Which means that $x=0$ is one of the roots and also means that the value of constant coefficient of $f$ is $0$.

And thats all that I've got and I'm not really sure what to do next.

Answer 1

Observation: If $n\geq 3$, $\max\{6,n+1\}<3n$

So you actually want the polynomial either:

1) has degree $0$ or $1$, then $f''(x)=0$, but then the LHS will be of degree $6$ and RHS will be of degree $3$, so it fails.

2) has degree $2$.
Then you can let $f(x)=Ax^2+Bx+C$ with $A,B,C\in \mathbb{Z}$ and the rest of the steps are left for you to calculate.

It ruins the fun...:

$$x^6+x^3\cdot 2A=Ax^6+Bx^3+C+x^3$$ By comparing coefficient of $x^2$ and comparing constant term, $$A=1, C=0$$ By substitution, $$x^6+2x^3=x^6+Bx^3+x^3$$ By comparing coefficient of $x$, $$B=1$$ So $f(x)=x^2+x$ is the only solution.