Proof - for all integers $y$, there is integer $x$ so that $x^3 + x = y$ For all integers $y$, there is an integer $x$ so that $$x^3 + x = y.$$
This is what I have done so far:
Proof: Suppose $y$ is some integer. We want to prove that $$x^3 + x = y$$ for some integer $x$.
I am thinking this is false. For example, if $y = 1$, there is no integer $x$ so that $x^3 + x = 1$. But I am not sure how to start about expressing this as a formal proof. Any help please?
 A: This is indeed not true. Two proofs:

Using basic calculus:
Note that $x^3+x$ is strictly increasing. (Its derivative is $3x^2+1$ and squares are positive)
Now note that $1^3+1=2$ and $0^3+0=0$. Therefore the $x$ such that $x^3+x=1$ must statistify $0<x<1$, so its not an integer.

Using number theory only:
Note that $x \mid x^3+x$, so $x^3+x=0$ or $|x^3+x|\geq |x|$. Thus it can only be 1, 0 or -1 if it is equal to one. But $x=-1$ gives $x^3+x=-2$, $x=0$ gives $x^3+x=0$ and $x=1$ gives $x^3+x=2$. So it is not possible for an integer.
A: Observe that $x^3+x=x(x^2+1)$ is a composite number if $x>1$, so can never represent a prime.
A: It depends on what you want to use. My first thought:
use the intermediate value theorem to show that $$f:\mathbb R\rightarrow \mathbb R,x\mapsto x^3+x-1$$ has a root between $0$ and $1$. Then show that $f$ is strictly increasing. So in this proof I use mainly some basic calculus.
On second thought, one could use, that for a polynominal $f(x)=x^n+\dots a_1x+a_0$ in $\mathbb Z[x]$ the following holds:
$$f(x_0)=0 \iff x_0\mid a_0.$$
Here we have $f(x)=x^3+x-1$, so $f(x_0)=0 \iff x_0=\pm 1$ and we can just pluck these two values in to see that $f$ doesn't have any (integer) roots.
Edit: of course there are other methods to proof this (see other answers), I mainly wanted to point out, that you don't have "the one proof" but depending on the context in which the problem arises, one usually has different approchaches (for a calculus class the first proof would be perfectly fine, for an algebra class not so).
A: How about this:
This is false.
Negation: There exists an integer $y$, such that for all integers $x$, $$x^3 + x \neq y.$$
Proof: Let $y$ = 1. Suppose $x^3 + x = y$, we will prove by contradiction. Then 
$$y = x^3 + x$$
$$1 = x^3 + x$$
$$1 = x(x^2 + 1),~\text{hence}~ x \neq 0$$
however, $x^2 + 1 > 1$ where $x \neq 0$, which contradicts with $1 = x(x^2 + 1)$. Therefore, $x \neq y$.
