# Sum of Squares Length of a Product

Let $n \geq 2$. Let $g_1, \ldots , g_{n-1} \in \mathbb{R}[x_1,\ldots,x_n]$ such that $q=g_1^2+\ldots +g_{n-1}^2$ is not divisible by $p=x_1^2+\ldots +x_n^2$. Let $m \geq 1$ be the smallest integer such that $p\cdot q$ is the sum of $m$ squares of polynomials. In the cases $n=2$ and $n=3$ it is rather easy to show that we always have $m \geq n$. Is this still true for $n >3$?

Edit: Using Pfister forms, one can show that it is true whenever $n=2^k+1$ for some $k \geq 0$.

The answer seems to be negative, already for $n=4$ and $n=8$.
For $n=1, 2, 4, 8$, we know there exists an identity
$$(x_1^2+\cdots+x_n^2)(y_1^2+\cdots+y_n^2)=z_1^2+\cdots+z_n^2,$$ where each $z_i$'s are bilinear on $x$'s and $y$'s, i.e., there exists an invertible rational matrix $P$ such that $z=x^tPy$, where $x^t=[x_1\cdots x_n], y^t=[y_1\cdots y_n], z^t=[z_1\cdots z_n]$. These identities comes from the norm of real, complex, Hamilton and Cayley numbers. Moreover one can always choose $z_1=x_1y_1+\cdots+x_ny_n$. Conversely Adolf Hurwitz proved that if such a identity happens then $n=1, 2, 4, 8$.
Now for your question choose the polynomials $g_1,\cdots,g_{n-1}\in\mathbb{R}[x_1,\cdots,x_n]$ such that $$x_1g_1+\cdots+x_{n-1}g_{n-1}=0, \ (*)$$ also put $g_n=0$. Using the above identity one obtains $$(x_1^2+\cdots+x_n^2)(g_1^2+\cdots+g_n^2)=z_2^2+\cdots+z_n^2.$$
Also for $n\geqslant 4$, in $(*)$ it is possible to choose $g_i$'s in such a way that $q=g_1^2+\cdots+g_{n-1}^2$ is not divisible by $p=x_1^2+\cdots+x_n^2$. So $m\leqslant n-1$.