Question:
Given a non-negative integer N, show many sets of non-negative integers (a,b,c,d) satisfy 2a+b+c+d=N
Proposed (and roadblocked) solution:
Case 1: 2a=0
Then there are \binom{N+2}{2} solutions (easy to prove).
Case 2: 2a=2
Then there are \binom{N+2-2}{2} solutions.
Case 3: 2a=4...