Basically, I'm looking for ways to multiply elements of $\mathbb{R}^n$ that allow me to count divisors in $\mathbb{Z}^n$.


For every positive integer $n$, I'm looking for an algebra structure on $\mathbb{R}^n$ such that 


1. Given $y,z \in \mathbb{Z}^n$ with $z$ non-zero, $x y = z$ has at most $c$ solutions $x \in \mathbb{Z}^n$, where $c$ is a constant independent of $y$ and $z$.
2. Given $z \in \mathbb{Z}^{n}$ with $z$ non-zero, $$\# \{ (x,y) \in \mathbb{Z}^{2n} : xy = z \} = \| z \|^{o(1)},$$ where $\| \cdot \|$ is the usual Euclidean norm on $\mathbb{R}^n$.


Can $\mathbb{R}^n$ be given such an algebra structure?


Examples: If $n = 2$, use $\mathbb{C}$. If $n=4$, maybe use the quaternions. It's not exactly clear what the divisor bound is for the quaternions.


I know that the only finite-dimensional real division algebras are the real numbers, the complex numbers, the quaternions, and the octonions.



Counting Divisors in $\mathbb{Z}^n$
I assume that conditions 1) and 2) are applies to nonzero $z$ only, otherwise each of them ia absurd.


Let me describe the construction more formally. Let $\circ$ be a componentwise multiplicaion on $\mathbb R^n$, i.e., $(x\_1,\dots,x\_n)\circ(y\_1,\dots,y\_n)=(x\_1y\_1,\dots,x\_ny\_n)$.


Choose a no-degenerate matrix $S$ such that each its column consists of $n$ algebraic numbers linearly independent over $\mathbb Q$. This ensures that for every nonzero $x\in\mathbb Z^n$ all the components of $xS$ are nonzero.


Now define
$$
 x\*y=e\bigl((xS)\circ(yS)\bigr)S^{-1},
$$
where $e$ is the base of natural logarithms.
We claim that $\*$ is a desired operation. One may easily see that the algebras $(\mathbb R,+,\*)$ and $(\mathbb R,+,\circ)$ are isomorphic, the isomorphism is provided by $x\mapsto exS$. Thus $\*$ is associative and distributive over addition.


Next, for every nonzero $x,y\in\mathbb Z^n$, the product $(xS)\circ(yS)$ consists of nonzero algebraic numbers, so $\bigl((xS)\circ(yS)\bigr)S^{-1}$ is a nonzero vector with algebraic components. Thus $x\*y$ contains at least one transcendent component and thus does not lie in $\mathbb Z^n$. Therefore, the conditions 1) and 2) are trivially satisfied (for nonzero $z$).


