Let ${\rm Harm}_n^d$ be the space of real harmonic polynomials in $n$ variables, homogeneous of degree $d$. If $P\in{\rm Harm}_n^d$, then $$\left(\frac{\partial^2}{\partial x_1^2}+\cdots+\frac{\partial^2}{\partial x_n^2}\right)P=0.$$ A harmonic polynomial is not necessarily irreducible in ${\mathbb R}[X_1,\ldots,X_n]$. For instance every non-zero $P\in{\rm Harm}_2^4$ splits as the product of two quadratic forms; it turns out that none of them is positive definite. Besides, it is not two difficult to show that if $X_1^2+\cdots+X_n^2$ divides a harmonic polynomial $P$, then $P=0$. These observations lead me to me following question:

Is it possible that a non-zero harmonic polynomial (say homogeneous) factorizes $P=QR$ in ${\mathbb R}[X_1,\ldots,X_n]$, with the factor $Q$ being positive definite (i.e. $Q(x)>0$ for every $x\ne0$) ?

I incline toward a negative answer, of course.

Let ${\rm Harm}_n^d$ be the space of real harmonic polynomials in $n$ variables, homogeneous of degree $d$. If $P\in{\rm Harm}_n^d$, then $$\left(\frac{\partial^2}{\partial x_1^2}+\cdots+\frac{\partial^2}{\partial x_n^2}\right)P=0.$$ A harmonic polynomial is not necessarily irreducible in ${\mathbb R}[X_1,\ldots,X_n]$. For instance every non-zero $P\in{\rm Harm}_2^4$ splits as the product of two quadratic forms; it turns out that none of them is positive definite. Besides, it is not two difficult to show that if $X_1^2+\cdots+X_n^2$ divides a harmonic polynomial $P$, then $P=0$. These observations lead me to me following question:

Is it possible that a non-zero harmonic polynomial (say homogeneous) factorizes $P=QR$ in ${\mathbb R}[X_1,\ldots,X_n]$, with the factor $Q$ being non-constant and positive definite (i.e. $Q(x)>0$ for every $x\ne0$) ?

I incline toward a negative answer, of course.