Solution
= Solution
Write $p=\sum_{j=0}^na_jx^{n-j}y^j$. Equating the coefficient of $x^{n-j-2}y^j$ in $\nabla^2p$ gives
$$
(n-j)(n-j-1)a_j+(j+2)(j+1)a_{j+2}=0.
$$
\b[Thus all even coefficients are determined by $a_0$ and all odd coefficients by $a_1$, with no further constraints. These two choices give two independent harmonic homogeneous <polynomials>.]