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.
 A: 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 $x=y$. 
A: On 1), what if $x = 1/2$ and $y = 0$?
What tools are you permitted to use?  Is calculus involved?
A: For (2), if $x^3+x=y^3+y$, then $x^3-y^3=y-x$. A standard factorization that you should learn (if you haven’t already) is $x^3-y^3=(x-y)(x^2+xy+y^2)$. If $x\ne y$, you can divide by $x-y$ to get
$$x^2+xy+y^2=\frac{y-x}{x-y}=-1\;.\tag{1}$$
Now write $(1)$ as a quadratic in $y$:
$$y^2+xy+(x^2+1)=0\;.$$
Use the quadratic formula to solve this for $y$ in terms of $x$; if you do it right, you’ll see why this is impossible (i.e., why there are no real solutions), and you’ll be able to conclude that $x-y$ must have been $0$ in the first place.
A: For 1) Observe that (I am assuming $x,y$ are real numbers):
\begin{align}
2x^2-x=2y^2-y\\
&\implies (2x^2-x)-2y^2+x=(2y^2-y)-2y^2+x\\
&\implies 2(x^2-y^2)=(x-y)\\
&\implies 2(x-y)(x+y)=(x-y)\\
&\implies 2(x-y)(x+y)-(x-y)=(x-y)-(x-y)=0\\
&\implies 2(x-y)(x+y)-(x-y)=0\\
&\implies (x-y)(2(x+y)-1)=0\\
&\implies (x-y)=0 \text{ or } 2(x+y)-1=0\\
\end{align} So, either $(x-y)=0$ or $2(x+y)-1=0.$ If $x-y=0$ then $x=y$ and we have our result. What if $2(x+y)-1=0$? Then there is a possiblity that $x\neq y.$ But if $x=y=1/4$ then $2(x+y)-1=2(1/4+1/4)-1=2(1/2)-1=1-1=0.$ Thus in either case $x=y$ is a solution. Like ncmathsadist pointed out $x=1/2$ and $y=0$ also works but that doesn't mean that $x=y$ doesn't work! I think for precalculus level this is enough. 
For 2) do a similar analysis and Brian has done enough for you in his answer.
A: For 2), $f'(x) = 3x^2 + 1 > 0$, so $f(x) = x^3 + x$ is strictly increasing on $\mathbb{R}$, and is therefore injective.
