Show that a multivariate polynomial is identically zero. 
Let $P(x_1,\cdots, x_n)$ be a polynomial with real coefficients in the variables $x_1,\cdots, x_n$ and suppose that $\left( \sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}\right)P(x_1,\cdots, x_n) = 0$ (identically) and that
$x_1^2+\cdots +x_n^2 $ divides $P(x_1,\cdots, x_n).$ Show that $P=0$ identically.

Let $Q =x_1^2 + \cdots +x_n^2,$ which is homogeneous of degree $2.$ We claim that since $Q$ is homogeneous, $P$ is divisible by $Q$ if and only if each of the homogeneous components of $P$ is divisible by $Q.$ If each of the homogeneous components of $P$ is divisible by $Q,$ then $P$ is clearly divisible by $Q.$ If $P$ is divisible by $Q,$ then $P = QR$ for some polynomial $R.$ Then for each homogeneous component $K_r$ of $R, QK_r$ is homogeneous, and forms a unique homogeneous component of $P.$ Thus each homogeneous component of $P$ is equal to $QK_r$ for some homogeneous component of $R,$ so each homogeneous component of $P$ is divisible by $Q.$ Thus, it is sufficient to consider the case where $P$ itself is homogeneous, say of degree $d.$
Suppose $P=Q^mR$ for some $m > 0,$ where $R$ is homogeneous of degree $d$ and not divisible by $Q$; note that homogeneity implies $\sum_{1\leq i\leq n}x_i\dfrac{\partial R}{\partial x_i} = dR.$ Indeed, let $R = K_1+\cdots + K_n,$ where each $K_i$ is a nonzero polynomial of degree $d.$ Then $\sum_{1\leq i\leq n}x_i\dfrac{\partial R}{\partial x_i} = \sum_{j=1}^n \sum_{1\leq i\leq n}x_i\frac{\partial K_j}{\partial x_i}\tag{1}.$
For each $1\leq i\leq n, $ let $p_i$ be the highest power of $x_i$ dividing $K_j$. Then $\forall 1\leq i\leq n, x_i\frac{\partial K_j}{\partial x_i} = x_i(p_ix_i^{p_i-1} K_{j ;i}),$ where $K_{j ; i}$ is the polynomial resulting from removing all occurrences of $x_i$ from $K_j.$ Then this equals $p_i x_i^{p_i} K_{j ; i} = p_i K_j,$ by the definition of $K_{j ; i}.$ Since $\deg(K_y) = d$ for all $y,\deg(K_j) = d,$ so by the definition of degree, $\sum_{1\leq i\leq n} p_i = d.$ Thus, $\sum_{1\leq i\leq n}x_i\dfrac{\partial K_j}{\partial x_i} = \sum_{1\leq i\leq n} p_i K_j = d K_j.$ Plugging this result into $(1)$ gives that $\sum_{1\leq i\leq n}x_i\frac{\partial R}{\partial x_i} = \sum_{i=1}^n d K_j = d R,$ by the definition of the $K_j$'s.
Now define $\nabla^2 $ by  $\nabla^2 := (\frac{\partial^2}{\partial x_1^2} + \cdots + \frac{\partial^2}{\partial x_n^2}).$ Then
\begin{align}0 &= \nabla^2 P =\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2} (Q^m R)\\
&= \sum_{i=1}^n \frac{\partial}{\partial x_i} 
 (\dfrac{\partial}{\partial x_i} (Q^m R)) \\
&= \sum_{i=1}^n \dfrac{\partial}{\partial x_i} \left((\dfrac{\partial}{\partial x_i} Q^m)R + Q^m(\dfrac{\partial}{\partial x_i} R)\right)\\
&= \sum_{i=1}^n \dfrac{\partial}{\partial x_i}\left((m Q^{m-1}2x_i)R + Q^m(\dfrac{\partial}{\partial x_i} R)\right) \\
&=\sum_{i=1}^n [m Q^{m-1}2 R + 2(2mx_i Q^{m-1} \frac{\partial}{\partial x_i} R + Q^m \dfrac{\partial^2}{\partial x_i^2} R] + \sum_{i=1}^n m(m-1)Q^{m-2} 4x_i^2 R\\
&= 2mnQ^{m-1}R + 4m d Q^{m-1} R + Q^m \nabla^2 R +   \sum_{i=1}^n m(m-1)Q^{m-2} 4x_i^2 R\\
&=  Q^m \nabla^2 R + (2mn + 4md )Q^{m-1} R +  4 m(m-1)Q^{m-1} R \end{align}

However, here I'm stuck as I'm not sure how to show that this forces $R$ to be divisible by $Q,$ a contradiction. Since $P$ was divisible by $Q$ by assumption, if $P$ is nonzero, then $m > 0$ (and is an integer), so this'll show that $P\neq 0$ is impossible and hence $P=0.$


Corollary 5.3 in $\textit {Harmonic Function Theory}$ (Axler, Bourdon and Ramey) is exactly the result you seek. See https://www.axler.net/HFT.pdf
