<s>[INST] Two proofs I'm having difficulty with I've been given an assignment. Almost done except the last two are tripping me up. They are as follows:
1) if $2x^2-x=2y^2-y$ then $x=y$
2) if $x^3+x=y^3+y$ then $x=y$
I imagine they use a similar tactic as they both involve powers, but I've tried factoring,completing the square, difference of squares and difference of cubes and nothing seems to help.
Any hints would be appreciated.
 [/INST] Question $1$: The equation is equivalent to $2x^2-2y^2=x-y$. The left-hand side factors as $2(x+y)(x-y)$. So our equation can be rewritten as
$$2(x+y)(x-y)=x-y.$$ 
Thus  any pair $(x,y)$ such that $x\ne y$ and $2(x+y)=1$ is a counterexample to the assertion that $x$ must be equal to $y$. This was pointed out by ncmathsadist. 
Edit: It turns out that one is supposed to show that the only integer solutions have $x=y$.  This follows from the above calculation, since $2(x+y)=1$ has no integer solutions.
Question $2$: We look at the question as amended. We can factor and rewrite the equation as $(x-y)(x^2+xy+y^2)=-(x-y)$. If $x\ne y$, we can cancel $x-y$, and obtain $x^2+xy+y^2=-1$. 
But the equation $x^2+xy+y^2=-1$ has no real solutions, since $x^2+xy+y^2\ge 0$ for all real $x$ and $y$.  One way to see this is to complete the square, getting 
$$x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2.$$
The right-hand side is  clearly never negative for real $x$ and $y$. So it is true that the only solutions of the original equation have $x=y$. 
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