$p \in \mathbb{R}[x_1, \dots, x_n]$ is divisible by $x_1^2+\dots+x_n^2$ and $\Delta p = 0$ where $\Delta = \frac{\partial}{\partial x_1} + \dots + \frac{\partial}{\partial x_n}.$ Prove that $p = 0.$
My solution is different from all the solutions I found online, which mostly involve reducing to the case where $p$ is homogeneous.
Proof: We will use the convention that $\mathbb{R}[x_k, x_{k+1}, \dots, x_n] = \mathbb{R}$ if $k > n.$ Write $p = (x_1^2 + \dots + x_n^2)q_0.$ If $q_0$ is not identically zero, then we can write $q_0 = x_1^{n_1}q_{n_1, 1} + x^{n_1 - 1}q_{n_1 -1 , 1} + \dots$ where $q_{i,1} \in \mathbb{R}[x_2, \dots, x_n], q_{n_1, 1} \ne 0.$ By direct calculation, $$0 = \Delta p = 2nq_0 + (\sum\limits_i x_i^2)\Delta q_0 + \sum\limits_i 4x_i \frac{\partial q_0}{\partial x_i}.$$ Comparing the coefficients of $x_1^{n_1+2}$ on each side, we get $0 = \Delta q_{n_1, 1}.$ Comparing the coefficients of $x_1^{n_1}$ on each side, we get $0 = 2nq_{n_1, 1} + \left(\sum\limits_{i > 1} x_i^2 \right) \Delta q_{n_1, 1} + n_1(n_1-1)q_{n_1, 1} + \sum\limits_{i > 1} 4x_i \frac{\partial q_{n_1, 1}}{\partial x_i} + 4n_1 q_{n_1, 1} = (2n+n_1^2 + 3n_1)q_{n_1, 1} + 4\sum\limits_{i > 1} x_i \frac{\partial q_{n_1, 1}}{\partial x_i}.$
Since $q_{n_1, 1} \ne 0,$ we may write $q_{n_1, 1} = x_2^{n_2} q_{n_2, 2} + \dots$ where $q_{i,2} \in \mathbb{R}[x_3, \dots, x_n]$ and $q_{n_2, 2} \ne 0.$ Comparing the coefficient of $x_2^{n_2},$ we get $0 = (2n+n_1^2+3n_1+4n_2)q_{n_2,2} + \sum\limits_{i>2} 4x_i \frac{\partial q_{n_2, 2}}{\partial x_i}.$ Defining $q_{n_k, k} \in \mathbb{R}[x_{k+1}, \dots, x_n]$ as the leading coefficient of $q_{n_{k-1}, k-1}$ and continuing this coefficient comparison process, we get $q_{n_k, k} \ne 0$ for all $k$ and $0 = (2n+n_1^2+3n_1+4n_2+\dots+4n_n)q_{n_n, n} + \sum\limits_{i>n} 4x_i \frac{\partial q_{n_n, n}}{\partial x_i} = cq_{n_n, n}$ where $q_{n_n, n} \ne 0 \in \mathbb{R}[x_{n+1}, \dots, x_n] = \mathbb{R}$ and $c = 2n+n_1^2+3n_1+4n_2+\dots+4n_n.$ But $c \ge 2n > 0,$ so $c \ne 0 \Rightarrow q_{n_n, n} = 0,$ contradiction. Thus, $q_0 = 0 \Rightarrow p = 0.$