So to prove a function is injective we assume that: $$\forall x,y \in \mathbb{Z},~f(x) = f(y) \to x = y$$
So this is what I've done thus far: $$x^{3} + 2x + 1 = y^{3} + 2y + 1$$ subtract 1 and factor $$x(x^{2} + 2) = y(y^{2} + 2)$$ and I know I want to show that $x = y$ and to do that here it seems I need to show the other factors are equivalent so that they vanish and I'm left with just $x = y$.
I just noticed that these factors $x^{2} + 2$ and $y^{2} + 2$ are $f'(x)$ and $f'(y)$ respectively. Could we say something like since we assume the functions are equal $(f(x) = f(y))$ then it follows that the derivatives are equivalent ($f'(x) = f'(y))$? Given that the function is continuous and smooth, of course.
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$\begingroup$You have: $f(x) = f(y) \implies x^3+2x + 1 = y^3 + 2y+ 1\implies (x-y)(x^2+xy+y^2 + 2)=0$. But $x^2+xy+y^2 + 2 = \left(x+\frac{y}{2}\right)^2+ \dfrac{3y^2}{4} + 2 > 0\implies x = y \implies $ one-to-one.
$\endgroup$ $\begingroup$The function is monotone increasing, and thus injective.
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