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.

  • 2
    $\begingroup$ Hint: take the derivative of $x^3 + x$ $\endgroup$
    – Zarrax
    Jul 18, 2012 at 5:13
  • 3
    $\begingroup$ Note: this has nothing to do with integers. For all reals $x,y$, if $x^3+x=y^3+y$, then $x=y$. $\endgroup$ Jul 18, 2012 at 5:15
  • 8
    $\begingroup$ However it's no longer true for complex numbers: $\mathrm{i}^3+\mathrm{i} = 0 = (-\mathrm{i})^3+(-\mathrm{i})$ $\endgroup$
    – celtschk
    Jul 18, 2012 at 10:13

8 Answers 8


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

  • 6
    $\begingroup$ The most elegant answer so far! +1 $\endgroup$
    – user22805
    Jul 18, 2012 at 7:35
  • 5
    $\begingroup$ @David Heffernan: No, I prefer it as it is -- short and to the point. $\endgroup$
    – TonyK
    Jul 18, 2012 at 9:39
  • 2
    $\begingroup$ @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. $\endgroup$ Jul 18, 2012 at 9:41
  • 7
    $\begingroup$ @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. $\endgroup$ Jul 18, 2012 at 9:46
  • 5
    $\begingroup$ 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 $xy^2>y^3$. $\square$ $\endgroup$
    – celtschk
    Jul 18, 2012 at 10:30

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.

  • $\begingroup$ 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! $\endgroup$
    – Brad
    Jul 18, 2012 at 7:03
  • 1
    $\begingroup$ @Brad How would you know that: ( x^2+y^2+xy+1 =/= 0) without doing the difference of two squares calculation? $\endgroup$
    – hayd
    Jul 18, 2012 at 9:27
  • 1
    $\begingroup$ @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 $ab\le 0$). $\endgroup$
    – celtschk
    Jul 18, 2012 at 10:10
  • 1
    $\begingroup$ @celtschk I wouldn't say that is any more obvious that the difference of two squares. $\endgroup$
    – hayd
    Jul 18, 2012 at 10:23
  • 2
    $\begingroup$ @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. $\endgroup$
    – celtschk
    Jul 18, 2012 at 10:34

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


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 $x=y$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – celtschk
    Jul 18, 2012 at 10:45
  • $\begingroup$ Yes, that is correct. $\endgroup$
    – Daryl
    Jul 18, 2012 at 21:15

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 $x, y$


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 $x=y$.

  • $\begingroup$ I don't think that x =/= y and x' =/= y' imply that (x+x') =/= (y+y')... $\endgroup$
    – b0fh
    Jul 18, 2012 at 9:22
  • $\begingroup$ 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 $x^3 + x \ne y^3 + y$. $\endgroup$ Jul 18, 2012 at 9:32
  • $\begingroup$ 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. $\endgroup$
    – hayd
    Jul 18, 2012 at 9:33
  • $\begingroup$ @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). $\endgroup$
    – hayd
    Jul 18, 2012 at 9:36
  • $\begingroup$ I thought it was clear enough I was not using that (which is clearly wrong). Will edit my answer accordingly. Thank you. $\endgroup$ Jul 18, 2012 at 9:44

Since $f(x) = x^3 + x$ is strictly monotone $f(x) = f(y) \implies x = y$.

  • $\begingroup$ This answer is completely contained in the answer by @copper.hat which was answered 7 hours before this one. $\endgroup$
    – GeoffDS
    Jul 18, 2012 at 21:28
  • $\begingroup$ Indeed, but mine is shorter and makes no unnecessary assumptions. Mathematics is about minimalism. $\endgroup$ Jul 27, 2012 at 13:29
  • 2
    $\begingroup$ What unnecessary assumptions did I make? Also, I provided two independent solutions :-). Mathematics is about understanding. $\endgroup$
    – copper.hat
    Jul 29, 2012 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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