$29 \cdot 53$ the difference of squares of two odd numbers is divisible by 8. If $a>b$, and $2 a+1$ and $2 b+1$ are two odd numbers, to prove the above statement, the difference of the squares of these numbers is
(A) $(2 a+1)^{2}-(2 b+1)^{2}$.
(B) $4 a^{2}-4 b^{2}+4 a-4 b$.
(C) $4[a(a+1)-b(b+1)]$.
(D) $4(a-b)(a+b+1)$.
(E) $4\left(a^{2}+a-b^{2}-b\right)$.
(5th American High School Mathematics Examination, 1954)
[Solution] From the given condition, note that $a>b$ implies
$$
\begin{aligned}
& (2 a+1)^{2}-(2 b+1)^{2} \\
= & 4 a^{2}+4 a+1-4 b^{2}-4 b-1 \\
= & 4[a(a+1)-b(b+1)],
\end{aligned}
$$
$a(a+1), b(b+1)$ are products of two consecutive integers, which must contain a factor of 2, so their difference is still even. Therefore, the answer is $(C)$.