Question 449516
If a and b are the integers, with *[tex |a| > |b|], then


*[tex \LARGE a^2 - b^2 = (a-b)(a+b)].


We can factor each of 2002, 2003, 2004, 2005. We must note that if there exist factors a-b and a+b with (a-b)(a+b) = 2002, ..., 2005, and the sum of these two factors is even (equal to 2a), then a and b are integers.


However, all factorizations of 2002 have an odd "sum" (e.g. 1*2002, 2*1001, 7*286, etc., in which their sums are 2003, 1003, 293). These numbers cannot be equal to twice an integer, so 2002 is the only impossible choice.