<s>[INST] 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?
 [/INST] 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.
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