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I need help proving the following statement:

For all integers $x$ and $y$, if $x^3+x = y^3+y$ then $x = y$

The statement is true, I just need to know the thought process, or a lead in the right direction. I think I might have to use a contradiction, but I don't know where to begin.

Any help would be much appreciated.

Martin Sleziak
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Brad
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8 Answers8

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If $x>y$ then $x^3 > y^3$ so $x^3+x > y^3+y.$

Ragib Zaman
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    The most elegant answer so far! +1 –  Jul 18 '12 at 07:35
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    +1 But it would be better if you included justification for the statement that $x>y$ implies $x^3>y^3$. – David Heffernan Jul 18 '12 at 08:25
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    @David Heffernan: No, I prefer it as it is -- short and to the point. – TonyK Jul 18 '12 at 09:39
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    @TonyK Well, you and I know why that key step in the proof is true, but perhaps the person asking the question does not know why. – David Heffernan Jul 18 '12 at 09:41
  • @Ragib Zaman To prove that $x>y\Rightarrow x^3>y^3$ you have to use the equality $x^3-y^3=(x-y)(x^2+y^2+xy)$. – blindman Jul 18 '12 at 09:42
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    @blindman That's a very strong statement. I doubt that the only way to prove that $f(x)=x^3$ is strictly monotone increasing is to use that equality. – David Heffernan Jul 18 '12 at 09:46
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    Indeed, you can proof it without that inequality. First, if $x=0$ or $y=0$, it is easy to proof by multiplying with $y^2$ resp. $x^2$. Next, for $x>0$ and $y<0$, we just proved that $x^3>0$ and $y^3<0$, thus $x^3>y^3$. In the case $x<0$ we can use that $(-x)^3=-x^3$ to show that the statement is equivalent to "if $-y>-x$, then $(-y)^3>(-x)^3$", reducing it to the case $y>0$. In that case, we can multiply the inequality with $x^2$ to find $x^3 > x^2y$. We can also multiply the inequality with $xy$ to find $x^2y>xy^2$. And finally we can multiply with $y^2$ to find $xy^2>y^3$. $\square$ – celtschk Jul 18 '12 at 10:30
  • Erm, the 5 minute edit threshold passed :-( I of course ment "without that equality" (the one for $x^3-y^3$). – celtschk Jul 18 '12 at 10:39
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    @celtschk: After you reduce to case $x,y>0$ you can just say $x^3 = x \cdot x \cdot x > y \cdot x \cdot x > y \cdot y \cdot x > y^3$, and in general $x>y>0 \implies x^n > y^n$. – sdcvvc Jul 18 '12 at 13:02
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    @sdcvvc: That's exactly what I did, except that I was explicit about how to get to the single inequalities in that chain. – celtschk Jul 18 '12 at 13:22
  • @Brad In case you haven't learned this sort of thing yet this is called a proof by the contrapositive. A priori you can assume wlog that $x > y$. –  Jul 21 '12 at 14:30
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We have \begin{eqnarray*} x^3+x=y^3+y&\Longleftrightarrow& (x^3-y^3)+(x-y)=0\\ &\Longleftrightarrow& (x-y)(x^2+y^2+xy+1)=0. \end{eqnarray*} Since $x^2+y^2+xy+1=(x+\frac{y}{2})^2+\frac{3}{4}y^2+1>0$, we get $x=y$. The hypothesis $x,y$ are integer numbers is redundant.

blindman
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  • You could even just say that (x^2+y^2+xy+1) =/= 0, and conclude x-y = 0, and thus x=y, correct? Anyway, thank you for your help! – Brad Jul 18 '12 at 07:03
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    @Brad How would you know that: ( x^2+y^2+xy+1 =/= 0) without doing the difference of two squares calculation? – hayd Jul 18 '12 at 09:27
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    @haydoni: Be $a=|x|$ and $b=-|y|$. It is obvious that $x^2+y^2+xy+1 \ge a^2+b^2+ab+1 \ge a^2+b^2+2ab+1 = (a+b)^2+1 > 0$ (the second inequality holds because by construction $ab\le 0$). – celtschk Jul 18 '12 at 10:10
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    @celtschk I wouldn't say that is any more obvious that the difference of two squares. – hayd Jul 18 '12 at 10:23
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    @haydoni: I didn't say it was more obvious (although to me it is, but that's obviously subjective). I just answered the question how you could know it without doing that difference. – celtschk Jul 18 '12 at 10:34
  • @celtschk haha, ok my bad, what I really meant to say was "how can you say that without doing some kind of calculation". – hayd Jul 18 '12 at 10:55
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The function $f(x) = x^3+x$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$. Hence $f(x)= f(y)$ iff $x=y$.

Since $f'(x) = 3x^2+1 \geq 1$, it is strictly increasing (hence injective), and since $\lim_{x\to -\infty} f(x) = -\infty$ and $\lim_{x\to +\infty} f(x) = +\infty$, its range is $\mathbb{R}$ (hence surjective).

Indeed, if you want a more tedious approach, you could check that $$g(x) = \frac{\sqrt[3]{\sqrt{27x^2+4}+3 \sqrt{3} x)}}{\sqrt[3]{2} \sqrt{3}} - \frac{\sqrt[3]{2}}{\sqrt{3} \sqrt[3]{\sqrt{27 x^2+4}+3 \sqrt{3} x}}$$ satisfies $(g \circ f) (x) = x$.

copper.hat
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Consider the following:

$$x^3+x=y^3+y\rightarrow x^3-y^3+x-y=0$$

This can be factored as $$(x-y)(x^2+xy+y^2+1) =0.$$

The problem is now reduced to factoring this expression over $x,y\in\mathbb{Z}$.

From the first factor, $x=y$.

From the second factor, for given $y\in\mathbb{R}$, $$x=\frac{-y\pm i\sqrt{3y^2+1}}{2}\in\mathbb{C}\backslash\mathbb{R}.$$

Thus, the only solutions $x,y\in\mathbb{Z}$ are $x=y$.

Daryl
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A slight variation on some of the other answers:

Rewrite as:

$x^3 - y^3 = -(x - y)$

Clearly this equation holds for $x = y$ but, to hold in any other case requires that:

$(x^3 - y^3) < 0 $ for some $x > y$ or $(x^3 - y^3) > 0 $ for some $y > x$

But, it's easy to show that this isn't the case for real $x, y$

Alfred Centauri
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If you want to use a contradiction:

If $x \ne y$ then $x^3+x \ne y^3+y$

Thus, $x=y$.

EDIT: this is true because $x^3+x$ is an increasing function (its derivative is $3x^2+1$, which is always > 0). Therefore:

if $x>y$, $f(x)>f(y)$ ==> $x^3+x$ > $y^3+y$

if $x<y$, $f(x)<f(y)$ ==> $x^3+x$ < $y^3+y$

Some said the "integer" requirement is redundant. Well, actually it isn't. It's just superabundant.

In fact, that statement is true only if $x$ and $y$ are real numbers (integers are included in the real field). If there were no restrictions, we would have to consider $x$ and $y$ as complex numbers, and this statement would not be true.

In the case of my proof, $x^3 + x=y^3 + y$ would not imply $x=y$ because of the Fundamental Theorem of Algebra. Considering $y$ as a constant in $x^3 + x=y^3 + y$ (which can be factored in $(x-y)(x^2+xy+y^2+1) = 0$), we would obtain 3 solutions. One being real ($x=y$) and two being complex conjugate:

$$x=\frac{-y\pm i\sqrt{3y^2+1}}{2}$$

You can see why, if $x$ and $y$ are complex, $x^3 + x=y^3 + y$ does not imply $x=y$.

Ivano.Bilenchi
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  • I don't think that x =/= y and x' =/= y' imply that (x+x') =/= (y+y')... – b0fh Jul 18 '12 at 09:22
  • Yep, that is not always true. But if $x' = x^3$ and $y' = y^3$:

    $x>y$ implies $x^3>y^3$, and $x^3 + x > y^3 + y$ //// $x<y$ implies $x^3<y^3$, and $x^3 + x < y^3 + y$ ////

    Therefore, $x \ne y$ implies $x^3 \ne y^3$, and $x^3 + x \ne y^3 + y$.

     – Ivano.Bilenchi Jul 18 '12 at 09:32
  • You're saying that $x=y$ doesn't imply $x^3=y^3$, i.e. cubed isn't function over the complex numbers?? You're wrong. – hayd Jul 18 '12 at 09:33
  • @Exile.90 You are implicitly using the fact the $f(x)=x^3+x$ is increasing, this needs to be demonstrated or at least mentioned. Your argument seems to use $a \neq b$ and $c \neq d$ $\implies$ a+c $\neq$ b+d (which is not true). – hayd Jul 18 '12 at 09:36
  • I thought it was clear enough I was not using that (which is clearly wrong). Will edit my answer accordingly. Thank you. – Ivano.Bilenchi Jul 18 '12 at 09:44
  • So you're not using "then $x^3 \neq y^3$ in line 2 at all, this was the confusing bit. I now see what you are trying to say in the last part, well done. – hayd Jul 18 '12 at 10:21
  • Yep, I understand it was very confusing. One shouldn't do maths right after he wakes up (:p). Edited for further clarity. – Ivano.Bilenchi Jul 18 '12 at 10:44
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Since $f(x) = x^3 + x$ is strictly monotone $f(x) = f(y) \implies x = y$.

August Karlstrom
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Since overkill appears to be acceptable here: The discriminant of the polynomial $x^3 + x - d$ is given by $-4 - 27d^2 < 0$. As a result, for every real value of $d$, the equation $x^3 + x - d = 0$ has exactly one real solution and two complex conjugate solutions. So if $a^3 + a = b^3 + b$ for real $a$ and $b$, then $a = b$, as they are both real solutions of $x^3 + x - d = 0$, where $d = a^3 + a = b^3 + b$.

Zarrax
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