For all integers $x$ and $y$, if $x^3 + x = y^3 + y$ then $x = y$.
This is what I have done so far:
Proof: Suppose $x$ and $y$ are arbitrary integers. We know that $x^3 + x = y^3 + y$, we want to prove that $x = y$.
So, this is logically making sense, so it is true. any hints please?
$$ x^3+x=y^3+y\Longrightarrow x^3 - y^3 + x - y = 0\Longrightarrow (x-y)(x^2 + xy + y^2 + 1) =0 $$ But $$ x^2 + xy + y^2 + 1 = \left(x+\frac y2\right)^2 + \frac34 y^2 + 1 > 0, $$ and $x=y$ immediately.
Actually, that follows from that: $f(x)=x^3+x$ is strictly increasing (because $f'(x)=3x^2 + 1 > 0$ for all $x$). Taking derivative is much faster.
Proof:
$x^3+x = y^3+y$
$x^3-y^3+x-y = 0$
$(x-y)(x^2+xy+y^2) + x-y = 0$
$(x-y)(x^2+xy+y^2+1) = 0$
Either$ x-y = 0$ or $(x^2+xy+y^2+1) = 0$
The reason why $(x^2+xy+y^2+1)$ is $((x^2+(\frac{y}{2})^2) + \frac{3}{4}y^2 + 1)$ where for any integer x and y, you have the two squares which can be 0 or greater than 0 + 1. Hence the term cannot be equal to 0. The other responder has made this point which was very obvious to me before.
But the latter is not equal to 0 for integers x and y
Hence $x-y = 0 \Longrightarrow x= y$
Note that $f(x) = x^3 + x$ is a strictly increasing function of $x$. As $x$ and $y$ increase, so too does $f(x)$ and $f(y)$.
Assume for contradiction there exists $y\neq x$ such that $f(x) = f(y)$. Then, WLOG, let $x > y$. We have $f(x) = f(y)$, but $f(x) = f(x)$, which means the function $f(\cdot)$ in $[y,x]$ has slope $0$. This is a contradiction to the fact that $f(x)$ is strictly increasing.
$$x^3-y^3+x-y=(x-y)(x^2+xy+y^2+1)=0$$ has no other solution than $x=y$, because solving the quadratic term for $y$ give a negative discrimiant $x^2-4(x^2+1)$.