1954 AMC 12 Problem 49

Question

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:

(2a+1)^2-(2b+1)^2

4a^2-4b^2+4a-4b

4[a(a+1)-b(b+1)]

4(a-b)(a+b+1)

4(a^2+a-b^2-b)

Expand the two squares and group each variable with its successor

Each product

a(a+1)

and

b(b+1)

is even

Expanding and regrouping,

\begin{aligned} &(2a+1)^2-(2b+1)^2\\ &\qquad=4[a(a+1)-b(b+1)]. \end{aligned}

Each of

a(a+1)

and

b(b+1)

is even, so their difference is even. The displayed expression is consequently divisible by

4\cdot2=8.

Thus, the correct answer is C.

Problem 49 in other years