SOLUTION: No matter what two integers I choose, their squares cannot differ by A) 2002 B) 2003 C) 2004 D) 2005 Please explain your reasoning. Thank you.

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Question 449516: No matter what two integers I choose, their squares cannot differ by
A) 2002
B) 2003
C) 2004
D) 2005
Please explain your reasoning. Thank you.
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
If a and b are the integers, with , then

.

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.