If the square of every element of a ring is in the center, must the ring be commutative? Let $R$ be a ring with identity such that the square of any element belongs to the center of $R$. Is it necessary true that $R$ is commutative?
(I can show that for any $x,y\in R$, $2(xy-yx) =0 $ but I cannot prove commutativity of $R$.)

Here's a counterexample.  Consider the $\mathbb{F}_2$-algebra $R$ generated by two elements $x,y$ modulo relations that $x^2=y^2=0$ and every word of length $3$ formed by $x$ and $y$ is $0$.  Explicitly, $R$ has $\{1,x,y,xy,yx\}$ as a basis and any product of basis elements that would give a word not in this set is $0$.  Since $xy\neq yx$, $R$ is not commutative.
I now claim that the square of every element of $R$ is central.  Indeed, for an element $r=a+bx+cy+dxy+eyx\in R$ ($a,b,c,d,e\in\mathbb{F}_2$) we have
$$r^2 = a^2 + bcxy + bcyx.$$ To show that such an element is central, it suffices to show that $xy+yx$ is central.  But this is trivial, since $xy+yx$ annihilates both $x$ and $y$ on both sides.
