How many positive integral solutions $(x,\,y)$ are there for the equation $\dfrac{1}{x+1}+\dfrac{1}{y}+\dfrac{1}{y(x+1)}=\dfrac{1}{2015}$?

[anemone]

Here is this week's POTW:

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How many positive integral solutions $(x,\,y)$ are there for the equation $\dfrac{1}{x+1}+\dfrac{1}{y}+\dfrac{1}{y(x+1)}=\dfrac{1}{2015}$?

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[anemone]

No one answered to this POTW. (Sadface) However, you can find the suggested solution below:

Spoiler

$\dfrac{1}{x+1}+\dfrac{1}{y}+\dfrac{1}{y(x+1)}=\dfrac{1}{2015}$ simplifies to $xy-2015x-2014y=2(2015)$ and therefore $(x-2014)(y-2015)-2014(2015)=2(2015)$. Make $(x-2014)(y-2015)$ as the subject and factorize the other side as simplest as possible we have $(x-2014)(y-2015)=2^5\cdot 3^2 \cdot 5 \cdot 7 \cdot 13 \cdot 31$.

Hence, the number of factors of this number is $(5+1)(2+1)(1+1)(1+1)(1+1)(1+1)=288$.