A problem by Karan Arora

$666\equiv2\pmod4$ , but $x^2$ and $y^2$ are each congruent to 0 or 1, so that $x^2-y^2$ is congruent to 0, 1, or 3. Thus there are no solutions.

We can write the set $S$ as the factored form $(x-y)(x+y) = 666 = 2^{1}3^{2}37^{1}$ , which can be grouped into the following positive integer divisor pairs:

$x+y = 666, 333, 222, 111, 74, 37$

$x-y = 1, 2, 3, 6, 9, 18$

which after adding these two subsets together gives:

$x = \frac{667}{2}, \frac{335}{2}, \frac{225}{2}, \frac{117}{2}, \frac{83}{2}, \frac{55}{2}$

Thus, $x,y \notin \mathbb{P} \Rightarrow \boxed{\mathbb{S} = \emptyset}$