We define $f: Z\to Z, f(x)= x^3-2x$ is f injective ? But surjective? We define $f: Z\to Z, f(x)= x^3-2x$ is f injective ? But surjective ?
I think f is injective and not surjective, but I'm having a hard time trying to prove it.$..f(3)=21,f(2)=4,f(1)=-1,f(0)=0,f(-1)=1,f(-2)=-4,f(-3)=-21...$.I think f is ascending on $Z-\{1,-1\}$. I can see that there is no $x\in Z$ such that $f(x)=2$.
Proof f is injection: Let $x_1,x_2\in Z$. Assume that $f(x_1)=f(x_2)$
$x_1^3-2x_1 = x_2^3-2x_2$. I'm a little bit stuck here.
Proof f is surjection: Assume f is surjection. For any $y\in Z$ there is a $x\in Z$ such that $f(x)=y$. How can I find a value for x that maps to y ? 
 A: $$f(x)=f(y)\iff x^3-2x=y^2-2y\iff(x-y)(x^2+xy+y^2)=2(x-y)$$
If $\;x=y\;$ we're done, otherwise we get after cancelation $\;x^2+xy+y^2-2=0\;$ . This is a quadratic in $\;x\;$  say, and its discriminant is
$$\Delta=y^2-4(y^2-2)=-3y^2+8$$
The above must be non-negative if there is a solution:
$$-3y^2+8\ge0\iff y^2\le\frac83\iff |y|\le\frac{2\sqrt2}{\sqrt3}\cong1.63$$
so it must be $\;y=-1,\,0,\,1\;$ , and we then get
$$x^2-2=0 \;\;\text{-- no integer solution -- , or}\;\;x^2\pm x-1=0$$
and again no integer solution,. so the function is $\;1-1\;$ .
It is not surjective since for example there is no integer $\;x\;$ such that $\;x^3-2x=5\;$ , since then
$$x(x^2-2)=5\iff x=1\;,\;\;x^2-2=5$$
or any other combination with signs, and this has no integer solutions.
A: $f'(x)=3x^2-2>0$ for $x\ge1$, and so $f$ is strictly increasing for $x\ge1$. But $f(2)=4$ and $f(3)=21$, for example, and so no integer value between $4$ and $21$ is attained.
Use a similar argument for injectivity. Note $f'(x)>0$ for $x\ne0$, and so you only need to check what happens for $x\in[-n,n]$ with $n$ such that $f(-n)<f(m)$ and $f(n)>f(m)$ for $m\in\mathbb Z\cap[-n+1,n-1]$. Notice that $n=2$ suffices.
