Sleepy was sitting in class hearing his teacher talk about fractions. He learns that

\[ ` \frac{x}{n} + \frac{y}{n} = \frac{ (x+y)} { n} \text{ because } x + y = (x+y) ' \]

before falling asleep. As punishment for this misdemeanor, his teacher made him solve \( \frac{1}{x} + \frac{1}{y} \) on the board. Drawing from the previous example, he said that \[ \frac{ 1}{x} + \frac{1}{y} = \frac{1}{(x+y)}. \]

How many ordered pairs of integers subject to \( -100 \leq x \leq 100, -100 \leq y \leq 100 \) are there, such that

\[ \frac{ 1}{x} + \frac{1}{y} = \frac{1}{(x+y)}? \]

×

Problem Loading...

Note Loading...

Set Loading...