Problem 1377

We have $2 \cdot 1 = (a+1)^2 - a^2 - 1 \in Z(\mathbb{K})$, for $a \in \mathbb{K}$. If the characteristic $char(\mathbb{K})$ of $\mathbb{K}$ is not $2$, then $2 \cdot 1$ is an invertible element in $Z(\mathbb{K})$, such that
$$a = (2 \cdot 1)^{-1} \cdot (2 \cdot a) \in Z(\mathbb{K}), \text{ for any } a \in \mathbb{K},$$
implying that $\mathbb{K}$ is abelian.

Consider the case $char(\mathbb{K}) = 2$. For $a, b \in \mathbb{K}$ we have $a = (a+b)^2 - a^2 - b^2 \in Z(\mathbb{K})$. As $w \cdot ab = (ab)^2 + ab^2a = (ab)^2 + a^2b^2 \in Z(\mathbb{K})$, if $w \neq 0$, then $a, b \neq 0$ and $ab = w^{-1} \cdot ((ab)^2 + a^2b^2) \in Z(\mathbb{K})$. Consequently $a \cdot (ab) = (ab) \cdot a = a \cdot (ba)$, so that, $ab = ba$.
In case $w = 0$, as $1 = -1$, we get $ab = -ba = ba$, that is $ab = ba$ for any $a, b \in \mathbb{K}$.