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

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.

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Hint: take the derivative of $x^3 + x$ – Zarrax Jul 18 '12 at 5:13
Note: this has nothing to do with integers. For all reals $x,y$, if $x^3+x=y^3+y$, then $x=y$. – Gerry Myerson Jul 18 '12 at 5:15
However it's no longer true for complex numbers: $\mathrm{i}^3+\mathrm{i} = 0 = (-\mathrm{i})^3+(-\mathrm{i})$ – celtschk Jul 18 '12 at 10:13

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.

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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 7:03
@Brad How would you know that: ( x^2+y^2+xy+1 =/= 0) without doing the difference of two squares calculation? – hayden Jul 18 '12 at 9:27
@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
@celtschk I wouldn't say that is any more obvious that the difference of two squares. – hayden Jul 18 '12 at 10:23
@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
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If $x>y$ then $x^3 > y^3$ so $x^3+x > y^3+y.$

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The most elegant answer so far! +1 – user22805 Jul 18 '12 at 7:35
@David Heffernan: No, I prefer it as it is -- short and to the point. – TonyK Jul 18 '12 at 9:39
@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 9:41
@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 9:46
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

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 $(g \circ f) (x) = x$.

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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 $x=y$.

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I guess instead of $\in\mathbb{C}$ you wanted to say $\in\mathbb{C}\setminus\mathbb{R}$ or $\notin\mathbb{R}$. All real numbers are in $\mathbb{C}$, too. – celtschk Jul 18 '12 at 10:45
Yes, that is correct. – Daryl Jul 18 '12 at 21:15

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 $x=y$.

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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 9:22 Yep, that is not always true. But if $x' = x^3$ and $y' = y^3$: $x>y$ implies $x^3>y^3$, and $x^3 + x > y^3 + y$ //// $x show 2 more comments A slight variation on some of the other answers: Rewrite as:$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$- Since$f(x) = x^3 + x$is strictly monotone$f(x) = f(y) \implies x = y$. - This answer is completely contained in the answer by @copper.hat which was answered 7 hours before this one. – Graphth Jul 18 '12 at 21:28 Indeed, but mine is shorter and makes no unnecessary assumptions. Mathematics is about minimalism. – August Karlstrom Jul 27 '12 at 13:29 What unnecessary assumptions did I make? Also, I provided two independent solutions :-). Mathematics is about understanding. – copper.hat Jul 29 '12 at 6:55 Since overkill appears to be acceptable here: The discriminant of the polynomial$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\$.

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