Exercise 9, § 3

Suppose given on E an (associative and commutative) addition under which all the elements of E are invertible and a multiplication which is non-associative, but commutative and doubly distributive with respect to addition. Suppose further that $n \in \mathbf{Z}, n \neq 0$ and $nx = 0$ imply $x = 0$ in E. Show that if, writing $[x, y, z] = (xy)z - x(yz)$, the identity

holds, then $x^{m+n} = x^mx^n$ for all $x$ (show, by induction on $p$, that the identity $[x^q, y, x^{p-q}] = 0$ holds for $1 \leq q < p$).