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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.

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1  
Is (2) correct as written, or should the righthand side be $y^3+y$? As written, it’s very easy, since it reduces to $x^3=y^3$, and the cubing function is one-to-one. –  Brian M. Scott Jul 12 '12 at 1:21
    
Yes that is correct. My bad. I should also say that I can also show a counterexample, but these seem true to me –  user979616 Jul 12 '12 at 1:40
    
If you move everything over to (say) the left-hand side of each equation, you can factor-out "$x-y$". If this factor can be presumed zero, then $x=y$. However, in each case, you have another factor that could be zero as well; you need to rule-out this possibility. What do your instructions tell you about $x$ and $y$? (For instance, are they integers?) [Ah, just saw your other comment] You can use these other factors to find counter-examples; just choose values that make them zero. –  Blue Jul 12 '12 at 1:50
    
Isn't it true that if f(x) =f(y), then x=y? –  jeff Jul 12 '12 at 3:59
    
@jeff: Consider $f$ the square function. We have $f(-2) = 4 = f(2)$, but $-2 \neq 2$. –  Blue Jul 12 '12 at 4:19

5 Answers 5

up vote 3 down vote accepted

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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Ah for part 1 I got that far. But since x and y have to be integars, there are no such integars such that x+y=1/2. So am I right to conclude that x=y –  user979616 Jul 12 '12 at 3:37
    
@user979616: Certainly the only integer solutions have $x=y$, since the equation $2(x+y)=1$ has no solution in integers. Your post made no mention that variables were restricted to be integers. –  André Nicolas Jul 12 '12 at 3:41

On 1), what if $x = 1/2$ and $y = 0$?

What tools are you permitted to use? Is calculus involved?

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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.

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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.

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For 2), $f'(x) = 3x^2 + 1 > 0$, so $f(x) = x^3 + x$ is strictly increasing on $\mathbb{R}$, and is therefore injective.

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