<s>[INST] 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 $x,y\in\mathbb{Z}.$
 [/INST] 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).
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