How to prove that if $2x^2-x=2y^2-y$, then $x=y$, for $x,y\in\mathbb{Z}.$ How to prove that if
$2x^2-x=2y^2-y$, then $x=y$, for $x,y\in\mathbb{Z}.$
 A: Subtract the two expressions, you get $$2x^2-x-(2y^2-y)=(2x^2-2y^2)-(x-y)=2(x-y)(x+y)-(x-y), $$ which factors as $(x-y)(2x+2y-1)$. Now notice that the second term $2x+2y-1$ is odd, so different from $0$, hence if the two expressions are equal, it must be the first term $x-y$ that is $0$. 
A: You want to show that the equation $z=2x^2-x$ has at most one integer solution. Recall the quadratic formula, which tells you that the solutions are
$$x=\frac{1}{4}\pm \frac{\sqrt{1+8z}}{4}.$$
You want to show that at most one of these can be an integer. But their sum is $\frac14$, which is not an integer, so obviously they cannot both be integers.
A: Let $y=x+k$ for some $k \in \mathbb{Z}^{\geq 0}$ (if $k<0$ then swap $x$ and $y$).  Then $$2x^2-x=2y^2-y$$ implies that
\begin{align*}
2x^2-x &= 2(x+k)^2-x+k \\
 &= (2x^2-x)+4kx+2k^2+k
\end{align*}
or equivalently that $k(4x+2k+1)=0$.  If $k \neq 0$, this implies $4x+2k+1=0$, and this gives a contradiction, since $4x+2k+1$ is odd and $0$ is even (like Andres Caicedo's answer).
A: $\displaystyle{x = y\quad\vee\quad x + y = {1 \over 2}.\quad}$ The second one is not possible. Then, $\displaystyle{\color{#ff0000}{\Large\quad x = y}}$ 
