Dimension of $V = \{P \in \mathbb{R}_n [X] : P (X^2)-( X^n + 1) P (X) =0\}$ 
For $n \in \mathbb {N}$, consider the vector space $V$ defined by
$$
V = \{P \in \mathbb{R}_n [X] : P (X^2)-( X^n + 1) P (X)  =0\}
$$
What is the dimension of $V$?

I know that $\forall P\in V\setminus\{0\}$ $\deg(P)=n$. Will this help me solve the problem?
 A: Here is another way to prove $\dim V = 1$. It is easy to find (guess) a solution $p(x)= x^n-1$ so $\dim V \geq 1$.
Now observe a transformation $\phi: V\to \mathbb {R}$ defined by $\phi (P):= P(2)$ which is clearly linear. Now let us show it is injective. Take $P\in {\rm ker}(\phi)$, then $\phi (P)=0$ so $P(2)=0$. But this means that $$P(4) = (2^n+1)P(2) =0$$
The same way we see (use formal induction) that $2^{2^n}$ is zero for $P$ for every $n\in \mathbb{N}$. But this means that $P(x)=0$ for all $x$ since it has infinite zeroes. So $\phi $ has trivial kernel and thus, by rank-nullity theorem $$\dim V = {\rm rang}(\phi) \leq  \dim \mathbb{R}=1$$
Conclusion $\dim V =1$.
A: Clearly $p(x)$ is not monomial, that is, exist second higest degree: $$p(x) = ax^n+bx^k+...$$ where $a\ne 0$ and $b\ne 0$. Obviously $k<n$. Then we get $$ax^{2n}+bx^{2k} +... = ax^{2n} +bx^{k+n} +ax^n+bx^k+...$$
By comparing the coefficients we get:

*

*If $k>0$: Then $b=0$ or $2k=k+n$ i.e. $k=n$, a contradiction.

*If $k=0$: Then $p(x) = ax^n+b$ so $$ax^{2n}+b = ax^{2n} +(a+b)x^{n} +b$$ so $b=-a$
So $V = \{a(x^n-1);a\in \mathbb{R}\} \implies \dim V =1$.
