If $P(x^{\beta})-P(x)=Q(x^{\alpha})-Q(x)$ then $P(x)=R(x^{\alpha})-R(x)$ Let $\alpha\geq2$ and $\beta\geq2$ two integers such that for every $a>0,\; b>0$ we have $\alpha^a\neq \beta^b.$ Let $P(x)=\sum_{i=1}^na_ix^i,\;a_i\in\mathbb{C}$ and $Q(x)=\sum_{i=1}^mb_ix^i,\;b_i\in\mathbb{C}$ such that
$$P(x^{\beta})-P(x)=Q(x^{\alpha})-Q(x).$$
Show that there exists $R(x)$ s.t. $P(x)=R(x^{\alpha})-R(x).$
@ Ewan Delanoy gives me a very nice idea (that is written below) but we have a problem with the proof that
$(W_2+W_3)\cap V_{\beta}=\lbrace 0 \rbrace.$
Can you help us?
 A: Since the OP is asking for hints, I intentionally do not provide a full solution. 
Quick hint : View $V={\mathbb C}[X]$ as a $\mathbb C$-vector space. The map $\phi_{\alpha} : V\to V, P \mapsto P(X^{\alpha})-P(X)$ is a linear endomorphism of $V$. Denote its image by $V_{\alpha}$. The hint is : compute a basis for $V_{\alpha}\cap V_{\beta}$.
If that hint is not enough, try 
More detailed hint :

 We can assume $\alpha<\beta$ (why?). Let $g=\gcd(\alpha,\beta)$ and $\alpha'=\frac{\alpha}{g},\beta'=\frac{\beta}{g}$. Next, define the following subsets of $V$ : ${\cal F}_1=\Bigg\lbrace \phi_{\alpha}(X^{\beta k}-X^{k}) \ \Bigg| \ k\geq 1\Bigg\rbrace, {\cal F}_2=\Bigg\lbrace \phi_{\alpha}(X^k) \ \Bigg| \ k\geq 1, \beta'\not| k \Bigg\rbrace$, ${\cal F}_3=\Bigg\lbrace \phi_{\alpha}(X^{\beta'k}) \ \Bigg| \ k\geq 1, g\not| k \Bigg\rbrace$. For each $i=1,2,3$, denote by $W_i$ the subspace spanned by ${\cal F}_i$. Show (in that order) that $V_{\alpha}=W_1+W_2+W_3$, $(W_2+W_3)\cap V_{\beta}=\lbrace 0 \rbrace, V_{\alpha}\cap V_{\beta}=W_1$. 

