For complex polynomials of many variables the following theorem holds:


If $p(x\_1,...,x\_n), q(x\_1,...,x\_n)\in \mathbb K[x\_1,...,x\_n]$, where $\mathbb K=\mathbb C$, are polynomials such that $p(x\_1,...,x\_n)$ is irreducible and for all $a\_1,...,a\_n \in \mathbb K$
$$
p(a\_1,...,a\_n)=0 \Rightarrow q(a\_1,...,a\_n)=0, 
$$
then $p(x\_1,...,x\_n) | q(x\_1,...,x\_n)$.


I look for a counterpart of this theorem in the case $\mathbb K=\mathbb R$ for polynomials of the second order. 
Some additional assumptions are necessary, because for example for $p(x\_1,x\_2)=x\_1^2+x\_2^2$ and $q(x\_1,x\_2)=2x\_1^2+x\_2$ the above theorem is not true.


Is it maybe true under additional assumption that $p,q$ are the second order and that


there exists a $y=(y\_1,...y\_n)\in \mathbb R^n$ such that $p(y)=0, grad f(y)\neq 0$ ?



About connection between divisibility and roots of real polynomials
Yes. If there is $y\in\mathbf R^n$ such that $p(y)=0$ and $\mathop{grad} p(y)\neq0$ then the set $\{p=0\}$ is a smooth real hypersurface of $\mathbf R^n$ in a neighborhood of the point $y$. This implies that $\{p=0\}$ is Zariski dense in the set $\{z\in\mathbf C^n\mid p(z)=0\}$. Hence, if $q$ vanishes on the real zero set $\{p=0\}$, it also vanishes on the complex zero set of $p$. It also implies that $p$ is irreducible in $\mathbf C[x\_1,\ldots,x\_n]$. Therefore, $p$ divides $q$ in $\mathbf C[x\_1,\ldots,x\_n]$ and hence in $\mathbf R[x\_1,\ldots,x\_n]$.


Note that the conditions on the degree of $p$ and $q$ are superfluous.


