Yes--I see how Case 2 fits in conceptually at least. I'm thinking about it as k=0 + 1 + 1 + \cdots + 1 + 0
and then the same argument for gaps between 1s, but for the gaps between a zero and a one, there are only 2 solutions (a, 0) and (-a, 0) for the 0+1+... gap and (0, -a) and (0, a) for the...
Prove that there are 4k solutions to the equation |a|+|b| = k, i.e. 4k pairs of a and b such that |a|+|b|=k.
The Attempt at a Solution
I've written out the first few:
for k = 0, the only solution is (0,0)
for k =1, we have (-1,0) (0, -1) (0,1)(1,0)...