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.
 A: 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 $x=y$.
A: If $x>y$ then $x^3 > y^3$ so $x^3+x > y^3+y.$
A: 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 $x, y$
A: 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.
A: 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 $x=y$.
A: Since $f(x) = x^3 + x$ is strictly monotone $f(x) = f(y) \implies x = y$.
A: 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 $(g \circ f) (x) = x$.
A: 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 $d = a^3 + a = b^3 + b$.
