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 $k \geq 0$.



Sum of Squares Length of a Product
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 $m\leqslant n-1$.


