Let $p\in \mathbb{C}[x_1,\dots,x_n]$ be an homogeneous polynomial with degree $d>0$, such that $\Delta p=0$, then $$ \frac{\partial p}{\partial x_i}x_j-\frac{\partial p}{\partial x_j}x_i\not\equiv 0 $$ for at least one $i\neq j$.
It looks simple, but I am stucked at it. Thanks in advance.
We first notice that proving the statement for real polynomials implies the result in complex case. Thus, I assume that $p \in \mathbb{R}[x_1, \ldots, x_n]$.
The statement means that $\nabla p(x)$ and $x$ are not collinear everywhere. Arguing by contradiction, assume $\nabla p(x) \wedge x$ vanishes everywhere. But $$0 = |\nabla p(x) \wedge x|^2 = \|\nabla p(x)\|^2\|x\|^2 - |\langle \nabla p(x), x \rangle|^2 = \|\nabla p(x)\|^2\|x\|^2 - d^2 |p(x)|^2$$ The last equality comes from the Euler's identity. In other words, $|p(x)| = \frac{1}{d} \|\nabla p(x)\| \|x\|$ for all $x \in \mathbb{R}^n$. But $p$ is a non-constant harmonic function, thus $p$ cannot reach a local maximum. Hence, $p(x) = -\frac{1}{d}\|\nabla p(x)\| \|x\|$ for all $x \in \mathbb{R}^n$. This is a contradiction with the fact that $p(0) = 0$ should be a global maximum.