Suppose that f: Z -> Z, with f(x) = 3x3 - x. Is f injective?
A function is injective if f(a) = f(b) => a = b.
Suppose that f(a) = f(b). Then:
3a³ - a = 3b³ - b b - a = 3(b - a)(b² + ab + a²)
So (b - a)[3(b² + ab + a²) - 1] = 0.
Here we have a = b as a solution, which would verify that f is injective over the integers. But I'm not convinced because of that quadratic in the other parentheses -- does that matter? Solving 3(b² + ab + a²) - 1 = 0 definitely doesn't lead to a = b...