F : Z → Z by f(n) = n^3 - 3n.

To show that f is injective, we need to show that if f(a) = f(b), then a = b for any a, b ∈ Z.

So, let f(a) = f(b). Then, we have:

a^3 - 3a = b^3 - 3b

Adding 3a and subtracting 3b from both sides, we get:

a^3 + 3a = b^3 + 3b

Now, we can factor both sides using the sum of cubes formula:

(a + 1)(a^2 - a + 1) = (b + 1)(b^2 - b + 1)

Since a, b ∈ Z, we know that a + 1 and b + 1 are also integers. Therefore, the only way for the above equation to hold is if both sides are equal and have the same sign.

If both sides are positive, then we have a + 1 = b + 1 and a^2 - a + 1 = b^2 - b + 1. Simplifying the first equation gives a = b, which is what we wanted to show.

If both sides are negative, then we have a + 1 = b + 1 and a^2 - a + 1 = b^2 - b + 1, but with opposite signs. Simplifying the first equation gives a = b, which is again what we wanted to show.

Therefore, f is injective.

To show that f is surjective, we need to show that for any z ∈ Z, there exists an n ∈ Z such that f(n) = z.

Let z ∈ Z be given. We want to find an n ∈ Z such that n^3 - 3n = z. We can rearrange this equation to get:

n^3 - 3n - z = 0

Now, we can use the rational root theorem to find a possible integer solution for n. Any integer solution must be a factor of z, so we look for factors of z in the form of ±1, ±2, ..., ±z. We can then test each possible factor to see if it satisfies the equation.

For example, if z = 27, then possible factors of z are ±1, ±3, ±9, ±27. Testing each of these factors, we find that n = 3 is a solution, since 3^3 - 3(3) = 27.

Therefore, for any z ∈ Z, there exists an n ∈ Z such that f(n) = z, and f is surjective.

Since f is both injective and surjective, it is bijective, and we can conclude that f : Z → Z is a bijection.