RMO Part-3!

$\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{2015}$

$\Rightarrow2015x+2015y=xy$

$\Rightarrow xy-2015x-2015y+\color{#3D99F6}{2015^{2}}=\color{#3D99F6}{2015^{2}}$

$\Rightarrow(x-2015)(y-2015)=\color{#3D99F6}{2015^{2}}$

Now, to find the number of positive pairs $(x,y)$ , we have to calculate the number of divisors of $\color{#3D99F6}{2015^{2}}$ ,

$\Rightarrow \color{#3D99F6}{2015^{2}}=5^{2}×13^{2}×31^{2}$

$\Rightarrow Divisors=3^{3}=\boxed{27}$

We have , $\frac{1}{x}$ + $\frac{1}{y}$ = $\frac{1}{2015}$

$\frac{1}{x}$ = $\frac{1}{2015}$ - $\frac{1}{y}$

$\frac{1}{x}$ = $\frac{y-2015}{2015×y}$

x= $\frac{2015y}{y-2015}$

x= $\frac{2015(y-2015)+2015^{2}}{y-2015}$

x=2015+ $\frac{2015^{2}}{y-2015}$

Now, for x to be an integer y-2015 should be a factor of $2015^{2}$ .

So , number of factors of $2015^{2}$ are 27. The answer is 27.

The number of positive integeral solution in $1/x + 1/y = 1/z$ is number of divisors of $z^2$ .

Using it here, no. of solutions are 27.

$\large \begin{array}{rl|l} \cfrac{1}{x}+\cfrac{1}{y}&=\cfrac{1}{2015}&\times2015xy\\ 2015y+2015x&=xy\\ 2015x+2015y-xy&=0&\text{Now we should factorise it}\\ 2015x-xy+2015y+{\color{#3D99F6}a}&={\color{#3D99F6}a}\\ x(2015-y)+2015y+{\color{#3D99F6}a}&={\color{#3D99F6}a}&\text{Now }{\color{#3D99F6}a}\text{ can be }-(2015\times2015)\text{ and }-(y\times y)\text{. Let's see the first.}\\ x(2015-y)-2015(-y+2015)&=-2015^2\\ (2015-y)(2015-x)&=2015^2 \end{array}$

The number of solutions is equal to the number of factors of $2015^2$ . $2015^2=5^2\times13^2\times31^2$ , so the solution is $(2+1)\times(2+1)\times(2+1)=27$