--description--

In the following equation x, y, and n are positive integers.

$$\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$$

For n = 4 there are exactly three distinct solutions:

$$\begin{align} & \frac{1}{5} + \frac{1}{20} = \frac{1}{4}\\ \\ & \frac{1}{6} + \frac{1}{12} = \frac{1}{4}\\ \\ & \frac{1}{8} + \frac{1}{8} = \frac{1}{4} \end{align}$$

What is the least value of n for which the number of distinct solutions exceeds one-thousand?

--hints--

diophantineOne() should return 180180.

assert.strictEqual(diophantineOne(), 180180);

--seed--

--seed-contents--

function diophantineOne() {

  return true;
}

diophantineOne();

--solutions--

// solution required