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

Solution 1:

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

Solution 2:

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.

Solution 3:

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

Solution 4:

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 $x=y$.