Ever heard about Egyptian Fractions?

$\text{The solution of the problem is equivalent of finding the unordered pairs} \\ \text{of (x, y) that satisfy the following equation: } \large{\frac1x + \frac1y = \frac{3}{2006}} \\ \Rightarrow \large{\frac{x + y}{xy} = \frac{3}{2006}} \\ \Rightarrow \large{\frac{3xy}{x+y} = 2006} \\ \Rightarrow \large{3xy = 2006x + 2006y} \\ \Rightarrow \large{3xy - 2006x = \frac{2006}{3}\cdot 3y} \\ \Rightarrow \large{x\cdot (3y - 2006) = \frac{2006}{3}\cdot (3y - 2006 + 2006)} \\ \Rightarrow \large{x\cdot (3y - 2006) = \frac{2006}{3}\cdot (3y - 2006) + \frac{2006^2}{3}} \\ \Rightarrow \large{(x - \frac{2006}{3})\cdot (3y - 2006) = \frac{2006^2}{3}} \\ \Rightarrow {(3x - 2006)\cdot(3y - 2006) = 2006^2} \\ \Rightarrow {(3x - 2006)\cdot(3y - 2006) = 2^2\cdot 17^2 \cdot 59^2} \\ \text{By just observing the equation we see that } 2 \equiv 17 \equiv 59 \equiv 2 \pmod {3} \\ \text{Since x and y are symmetrical in this equation we are looking solutions} \\ \text{to the equation } 2^a \cdot 17^b \cdot 59^c + 2006 \equiv 0 \pmod 3 , 0 \leq a, b, c \leq 2 \\ \Rightarrow 2^a \cdot 2^b \cdot 2^c + 2 \equiv 0 \pmod {3} \\ \Rightarrow 2^{a+b+c} + 2 \equiv 0 \pmod {3} \text{ Thus:} (a + b + c) = \Big\{0, 2, 4, 6\Big\} \\ \text{But since we are looking for unordered pairs, we can limit ourselves} \\ \text{such that: } (a + b + c) = \Big\{0, 2\Big\} \\ - \text{For (a + b + c) = 0 we get only 1 solution where } (a, b, c) = (0, 0, 0) \\ - \text{For (a + b + c) = 2 we get 6 solutions where } \\ (a, b, c) = (1, 1, 0), (1, 0, 1), (0, 1, 1), (2, 0, 0), (0, 2, 0), (0, 0, 2) \\ \text{Hence there are in total 7 solutions.}$