SOLUTION: prove that for all integers a, if a^3 is even, then a is even

An integer of the form 2n is even (where n is an integer).
An integer of the form 2n-1 is odd (where n is an integer).

The proof is by contradiction:

We assume that a³ is even, and a is odd.

Then a = 2k-1 and 

a³ = (2k-1)³ = (2k)³ + 3(2k)²(-1) + 3(2k)(-1)² + (-1)³ =

8k³ - 3(4k²) + 6k - 1 =

8k³ - 12k² + 6k - 1 =

Factor 2 out of the first three terms:

2(4k³ - 6k² + 3k) - 1

This is of the form 2n-1, which contradicts the
assumption that a is odd.

Edwin