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

  • Thread starter Thread starter anemone
  • Start date Start date
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Here is this week's POTW:

-----

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

-----

 
Physics news on Phys.org
No one answered to this POTW. (Sadface) However, you can find the suggested solution below:
$\dfrac{1}{x+1}+\dfrac{1}{y}+\dfrac{1}{y(x+1)}=\dfrac{1}{2015}$ 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 $(5+1)(2+1)(1+1)(1+1)(1+1)(1+1)=288$.
 
Back
Top