Find all integers x for which there exists an integer y such that 1/x + 1/y = \(\frac{1}{17}\)
\(\frac{1}{x} + \frac{1}{y} = \frac{1}{17}\)
\(\frac{x + y}{xy} = \frac{1}{17}\)
\(17(x + y) = xy\)
\(0 = xy - 17x - 17y\)
\(289 = (x - 17)(y - 17)\)
x-17 | y-17 | x | y |
1 | 289 | 18 | 306 |
17 | 17 | 34 | 34 |
289 | 1 | 306 | 18 |
The possible values of x are 18, 34, and 306.