problem: The positive difference between the squares of two consecutive odd integers is 128. What is the product of the two integers?
solution: Let $a, a+2$ be the two integers. We are given that $(a+2)^2-a^2 = 128$. Using the difference of squares factorization, the equation becomes $(a+2+a)(a+2-a) = 128$. Simplifying and solving, we get: \begin{align*}
  (2a+2)(2) &= 128\\
  \Rightarrow 2a+2 &= 64\\
  \Rightarrow 2a &= 62\\
  \Rightarrow a &= 31.\\
  \end{align*} Therefore the desired product is $a(a+2) = 31\cdot33 = \boxed{1023}$.