Proof that Laplacian is surjective $\mathcal{P}^n\to\mathcal{P}^{n-2}$ Let $\mathcal{P}^n$ denote the vector space of homogeneous polynomials on $\mathbb{R}^3$ of degree $n$. I need to prove that $\Delta|_{\mathcal{P}^n}:\mathcal{P}_n\to\mathcal{P}_{n-2}$, for $n\geq2$ is surjective, where $\Delta$ is the Laplace operator. 
The hint says that the proof should be done by inductive argument: in the inductive step I should conclude from the formula $$\Delta(x_1^{q_1}x_2^{q_2}x_3^{q_3})=q_1(q_1-1)x_1^{q_1-2}x_2^{q_2}x_3^{q_3}+q_2(q_2-1)x_1^{q_1}x_2^{q_2-2}x_3^{q_3}+q_3(q_3-1)x_1^{q_1}x_2^{q_2}x_3^{q_3-2}$$
that surjectivity of $\Delta|_{\mathcal{P}^n}:\mathcal{P}_n\to\mathcal{P}_{n-2}$ implies surjectivity of $\Delta|_{\mathcal{P}^{n+2}}:\mathcal{P}_{n+2}\to\mathcal{P}_n$.
I've tried things and things and I simply don't see how to do this. Help!
 A: This was a neat problem, thanks for asking. Since it's homework, I'm going to skimp on some details. First check that: $$\Delta(pq) = q \Delta p + p \Delta q + 2\sum_i \partial_i p \partial_i q$$
Let's show that $\Delta: \mathcal{P}^{n+2} \rightarrow \mathcal{P}^{n}$ is a surjection. It's manifestly linear, so it suffices to see each monomial $x^iy^jz^k$ is in the image, for $i+j+k = n$. Let's strongly induct downwards on $\max\{i,j,k\}$. The base case of $n$ is clear. Suppose we know we can get all of them with $\max\{i,j,k\} \geqslant M$, let's get $M-1$. 
WLOG we want to obtain the monomial $x^{M-1} y^- z^-$ (i.e. $x$ may as well be the variable with the largest power). Well, $\Delta$ is sorta like twice differentiating, so let's try $\Delta(x^{M+1} y^- z^-)$. From the above product rule, with $p = x^{M+1}, q = y^-z^-$, and ignoring constants, we have this equals $$\Delta(x^{M+1} y^- z^-) = x^{M-1} y^- z^- + x^{M+1}\Delta(y^- z^-) $$(check that $2 \sum_i \partial_i p \partial_i q$ vanishes). By inductive hypothesis, the second term lies in the image of $\Delta$ ($x$ appears to a high enough power), hence so does $\Delta(x^{M+1} y^- z^-) - x^{M+1} \Delta(y^- z^-)$, as desired. 
