Egyptian Fraction with New Years 2015

Rewrite the equation as

$\dfrac{x + y}{xy} = \dfrac{1}{2015} \Longrightarrow xy - 2015x - 2015y = 0$

$\Longrightarrow (x - 2015)(y - 2015) = 2015^{2}$ .

Now $2015 = 5*13*31$ , and so $2015^{2} = 5^{2}*13^{2}*31^{2}$ , which has $(2 + 1)(2 + 1)(2 + 1) = 27$ divisors. Now each of these divisors corresponds to a distinct solution pair $((x - 2015), (y - 2015))$ , which in turn corresponds to a distinct solution pair $(x,y)$ . Thus there are $\boxed{27}$ solution pairs.

Use Simon's Favorite Factoring Trick

$(x-2015)(y-2015)=2015^2$

As $2015^2=31^2\times13^2\times5^2 \therefore$ number of divisors of $2015$ is $3\times3\times3=\boxed{27}$