If a + b is even, then either both a and b are even or both are odd.
a could be written 2n or 2n+1, 2n1 (even or odd, odd)
b could be written 2k or 2k+1, 2k1 (even or odd, odd)
Substitute the even/odd representations in for a and b:
Even case:
a = 2n+3 = 2n+2+1 = 2(n+1) +1 which is odd
b = 2k1 which is odd
Odd case:
a = (2n+1)+3 = 2n+4 = 2(n+2) which is even
b = (2k+1)1 = 2k which is even
As the loop continues to add 3 to a each iteration it will eventually reach >=100 and run its course.
If a starts as any negative integer, it will not run.
I hope this helps.
2/3/2014

David M.