$x^{n(x,y)} - x$ commutes with $y$, then $R$ is commutative Theorem.
If in a ring $R$ for every pair of elements $x$ and $y$ we can find an
integer $n(x, y) > 1$ which depends on $x$ and $y$ so that $x^{n(x,y)}-x$ commutes with $y$,
then $R$ is commutative.
Proof.
Let $T$ be the subring of $R$ generated by $x$ and $y$. Suppose $t\in T$.
Thus for some integer $m > 1, t_1= t^m - t$ commutes with $x$. For some other
integer $n > 1, t_{2}=t_{1}^{n}-t_{1}$ commutes with $y$. Since $t_1$ commutes with $x$, $t_2$
also commutes with $x$. Thus $t_2$ commutes with both $x$ and $y$, and so with every
element in the subring they generate. Thus $t_2$ is in the centre of $T$. However
$t_2 = t_{1}^{n}-t_{1} = (t^m - t)^n - (t^m - t) = - (t^2p(t) - t)$ where $p(t)$ is a polynomial with integer coefficient. 
My question is why the last equation is equal to $ - (t^2p(t) - t)$ ?

Since $m$ and $n$ are both greater than $1$, the expression $f(t)=(t^m-t)^n-t^m$ is a polynomial in $t$ of degree $nm\ge 4$.  Also, notice that the term of lowest degree in $f(t)$ is either $t^n$ or $t^m$, which in both cases have degree $2$ or greater.  It follows that we can factor out a $-t^2$ from $f(t)$ to give $f(t)=-t^2p(t)$.
