# Difference between revisions of "1954 AHSME Problems/Problem 49"

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:

$\textbf{(A)}\ (2a+1)^2-(2b+1)^2\\ \textbf{(B)}\ 4a^2-4b^2+4a-4b\\ \textbf{(C)}\ 4[a(a+1)-b(b+1)]\\ \textbf{(D)}\ 4(a-b)(a+b+1)\\ \textbf{(E)}\ 4(a^2+a-b^2-b)$

## Solution

Although all of the forms listed can be used to show that the difference of $(2a+1)^2$ and $(2b+1)^2$ is necessarily divisible by $8$, we should use $\boxed{\textbf{(C)}}$ because at least one of the second and third factors are necessarily of different parities, so that their product is necessarily even and we are done.