I won't do the proof for you but I'll help you get the idea of how
to do it.
It may be enough to tell you that every positive even integer is
either of the form 4n+4 or -4n+2, for some integer -- if not, maybe
this discussion will help:
Make these sequences and correspondences between them. Notice how
they're split into sub-sequences left right and middle. Notice their
respective nth terms. We let 0 of the top sequence correspond to
0 of the multiples of 3 and to 2 of the positive multiples of 2.
{n} ..., -3, -2, -1, 0, 1, 2, 3, ... {n}
||| | | | | | | | |||
{3n} ..., -9, -6, -3, 0, 3, 6, 9, ... {3n}
||| | | | | | | | |||
{-4n+2} ..., 14, 10, 6, 2, 4, 8, 12, ... {4n+4}
The top sequence is the sequence of integers
The middle sequence is the sequence of integers divisible by 3.
The bottom sequence are the even positive integers, but they're
written according to this pattern:
We put 2 in the middle (corresponding to 0), then 4 on the right
of 2, then 6 on the left of 2, then 8 on the right of 4, then 10
on the left of 6, etc.
The sequence on the bottom right has nth term 4n+4, whereas the
sequence on the left has nth term -4n+2. Every positive even
integer is either of the form 4n+4 or -4n+2, for some integer.
You now show that the mapping is a bijection (into and onto).
Now you have enough equipment to write up a proof.
Edwin
