1995 AIME Problems/Problem 8

Problem

For how many ordered pairs of positive integers $(x,y),$ with $y<x\le 100,$ are both $\frac xy$ and $\frac{x+1}{y+1}$ integers?

Thus, for a given value of $y$ , we need the number of multiples of $y(y+1)$ from $0$ to $100-y$ (as $x \le 100$ ). It follows that there are $\left\lfloor\frac{100-y}{y(y+1)} \right\rfloor$ satisfactory positive integers for all integers $y \le 100$ . The answer is

^ Another way of stating this is to note that if $\frac{x}{y}$ and $\frac{x+1}{y+1}$ are integers, then $\frac{x}{y} - 1 = \frac{x-y}{y}$ and $\frac{x+1}{y+1} - 1 = \frac{x-y}{y+1}$ must be integers. Since $y$ and $y+1$ cannot share common prime factors, it follows that $\frac{x-y}{y(y+1)}$ must also be an integer.