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)
for k =2, we have (1,1) (-1, -1)(-1,1)(1,-1) (0,2)(2,0)(-2,0)(0,-2)
for k =3, we have (2,1)(1,2)(-2,1)(1,-2)(2,-1)(-1,2)(-1,-2)(-2,-1)(0,3)(3,0)(0,-3)(-3,0)
The pattern seems to involve looking at the order/sign permutations of the pairs. The total number of solutions seems to be:
4(for the permutations made by sign changes and flipping the order of the (0,k) pair) + n*8(where n is the number of pairs where a doesn't equal b and neither a nor b equals 0, and 8 is the number of permutations made by sign changes and flipping the order)+4 (if there is a pair where a=b (i.e. if k is even) for the sign/ordering permutations).
Then I would need to prove that n=(k-1)/2 rounded down, and prove k=0 and k=1 separately. I think proving what n equals could be accomplished by arguing that the first pair (in which the two elements are not equal) must be either ((k/2)+1, (k/2)-1) (if k is even) or ((k-1)/2, (k+1)/2) (if k is odd), and that there are only k/2 rounded down more pairs.
It seems like there should be an easier way of doing this, but I'm stuck at the moment. Any suggestions?