Pfister's 16-Square Identity and the norm of sedenions

Question

Consider the sequence of numbers: complex numbers $\Bbb C$, quaternions $\Bbb H$, octonions $\Bbb O$, and sedenions $\Bbb S$.

The Brahmagupta-Fibonacci 2-Square identity implies that the norm of the product of two complex numbers $a,b$ equals the product of their norms,

$$||ab|| = ||a||\,||b||\tag{1}$$

Analogous identities (Euler's 4-Square and Degen's 8-Square) do the same for $\Bbb H$ and $\Bbb O$.

Question: Does Pfister's 16-Square Identity imply that two sedenions $a,b$ will obey $(1)$ as well?

Answer 1

No. Note that sedenions have zero divisors, so the norm cannot be multiplicative. (Or it takes $0$ on non-zero sedenions, but that's just wrong).

Example: $(e_3 + e_{10})(e_6-e_{15}) = e_3 (e_6-e_{15}) + e_{10} (e_6-e_{15}) = (e_5 - e_{12}) + (e_{12} - e_5) = 0$. The norms of both $e_3 + e_{10}$ and $e_6-e_{15}$ are $2$.