How to prove that if $2x^2-x=2y^2-y$, then $x=y$, for $x,y\in\mathbb{Z}.$

Question

How to prove that if

$2x^2-x=2y^2-y$, then $x=y$, for $x,y\in\mathbb{Z}.$

I edited the original post. Is it the meaning you had intended? — Sujaan Kunalan, Oct 04 '13 at 01:21
From the original post, it seems more likely that OP meant $2x^2 - x = 2y^2 - y \implies x = y$. — manthanomen, Oct 04 '13 at 01:25
Agreed. This claim with the subtraction in the exponent is false: x=2, y=1... — Frank, Oct 04 '13 at 01:26

score 9 · Answer 1 · answered Oct 04 '13 at 01:32

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 $0$.

score 2 · Answer 2 · answered Oct 04 '13 at 01:32

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.

score 2 · Answer 3 · answered Oct 04 '13 at 01:44

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).

score 0 · Answer 4 · answered Oct 04 '13 at 01:45

0

$\displaystyle{x = y\quad\vee\quad x + y = {1 \over 2}.\quad}$ The second one is not possible. Then, $\displaystyle{\color{#ff0000}{\Large\quad x = y}}$

answered Oct 04 '13 at 01:45

Felix Marin

89,464

How to prove that if $2x^2-x=2y^2-y$, then $x=y$, for $x,y\in\mathbb{Z}.$

4 Answers4