Problem: The difference of the squares of two odd numbers is always divisible by 8. If a>b, and 2a+1 and 2b+1 are the odd numbers, to prove the given statement we put the difference of the squares in the form:
Answer Choices:
A. (2a+1)2−(2b+1)2
B. 4a2−4b2+4a−4b
C. 4[a(a+1)−b(b+1)]
D. 4(a−b)(a+b+1)
E. 4(a2+a−b2−b)
Solution:
(2a+1)2−(2b+1)2=4a2+4a+1−4b2−4b−1
=4[a(a+1)−b(b+1)]
Since the product of two consecutive integers is divisible by 2, the last expression is divisible by 8.