Difference of Squares
Source:
February 8, 2009
Problem Statement
The difference of the squares of two odd numbers is always divisible by . If , and 2a\plus{}1 and 2b\plus{}1 are the odd numbers, to prove the given statement we put the difference of the squares in the form:
(A)\ (2a\plus{}1)^2\minus{}(2b\plus{}1)^2 \\
(B)\ 4a^2\minus{}4b^2\plus{}4a\minus{}4b \\
(C)\ 4[a(a\plus{}1)\minus{}b(b\plus{}1)] \\
(D)\ 4(a\minus{}b)(a\plus{}b\plus{}1) \\
(E)\ 4(a^2\plus{}a\minus{}b^2\minus{}b)