I see, that simplifies the problem quite a bit -- thanks!

Nehushtan

2013-11-11 05:28:37

You have to determine whether there exist integers a and b such that and . If so, then the function defined is not injective. If not, then it is.

zetafunc.

2013-11-11 04:08:53

Suppose that f: Z -> Z, with f(x) = 3x^{3} - 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...