It is well-known that the product of two squares is a square, and the product of sums of two squares is a sum of two squares (Brahmagupta-Fibonacci identity): $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$$ and an analogous result holds for sums of four squares (Euler's four square identity), eight squares (Degen's eight square identity) and 16 squares (Pfister's 16 square identity).
According to the Wikipedia article, analogous results hold for all sums of $2^n$ squares. But what about sums of $2n$ squares? Is it known whether there is a formula for the product of sums of 6 squares as a sum of 6 squares, i.e. whether, given $a_i,b_j\in\mathbb{Z}$, $$\left(\sum_{i=1}^6a_i^2\right)\left(\sum_{j=1}^6b_j^2\right)=\sum_{k=1}^6c_k^2$$ where $c_k=f_k(a_1,\ldots,a_6,b_1,\ldots,b_6)$ for some polynomials $f_k$?
There is no such formula.
Pfister’s theorem tells us that in any field, when $n$ is a power of 2, the set of sums of $n$ squares is closed under multiplication. It follows that when $n$ is a power of 2, there is a formula for the $c_k$ in terms of $a_i,b_i$, where $(a_1^2+\ldots+a_n^2)(b_1^2+\ldots+b_n^2)=c_1^2+\ldots+c_n^2$ .
Another theorem of Pfister (converse of Pfister’s above theorem) tells us that for every positive integer $n$ that is not a power of 2, the nonzero sums of $n$ squares are not closed under multiplication in the field $\mathbb{R}(x_1,x_2,\ldots)$. This means that there are two sums of $n$ squares in this field such that their product is not a sum of $n$ squares in the field. Hence no formula exists when $n$ is not a power of 2.
Source: https://kconrad.math.uconn.edu/blurbs/linmultialg/pfister.pdf