74th Romanian Mathematical Olympiad

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 $w\cdot ab=(ab{)}^2+a{b}^{2}a=(ab{)}^2+a^2{b}^2\in Z(\mathbb{K})$, if $w\ne 0$, then $a,b\ne 0$ and $ab=w^{-1}\cdot ((ab{)}^2+a^2{b}^2)\in Z(\mathbb{K})$. 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 $a,b\in \mathbb{K}$.