## Integral Points

A point $(x,y)$ is called integral if both $x$ and $y$ are integers. How many points on the graph of $1/x+1/y=1/4$ are integral points?
Source: NCTM Mathematics Teacher, November 2006

SOLUTION
Divide $1/4$ by $2$
$1/4=1/8+1/8$, $(x,y)=(8,8)$
Divide by $3$
$1/4=1/12+2/12=1/12+1/6, (x,y)=(6,12),(12,6)$
Divide by $4$
$1/4=1/16+3/16$, no solution
$1/4=2/16+2/16=1/8+1/8$, duplicate solution $(8,8)$
Divide by $5$
$1/4=1/20+4/20=1/20+1/5, (x,y)=(5,20),(20,5)$
$1/4=2/20+3/20$, no solution
Divide by $6$
$1/4=1/24+5/24$, no solution
$1/4=2/24+4/24=1/12+1/6$, duplicate solution $(6,12)$
$1/4=3/24+3/24=1/8+1/8$, duplicate solution $(8,8)$
Divide by $7$
$1/4=1/28+6/28$, no solution
$1/4=2/28+5/28$, no solution
$1/4=3/28+4/28$, no solution
Divide by $8$
$1/4=1/32+7/32$, no solution
$1/4=2/32+6/32$, no solution
$1/4=3/32+5/32$, no solution
$1/4=4/32+4/32=1/8+1/8$, duplicate solution $(8,8)$
We found five solutions $(8,8),(6,12),(12,6),(5,20),(20,5)$. We need an algebraic way to find all the solutions.
$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{4}$
$\dfrac{x+y}{xy}=\dfrac{1}{4}$
$xy=4(x+y)$
$xy=4x+4y$
$xy-4y=4x$
$y(x-4)=4x$
$y=\dfrac{4x}{x-4}=4+\dfrac{16}{x-4}$
For $y$ to be an integer $x-4$ must divide $16$. It’s positive divisors $1,2,4,8,\!16$ yield the five previous solutions. The negative divisors $\textrm{-}1,\textrm{-}2,\textrm{-}4,\textrm{-}8,\textrm{-}16$ yield four more solutions, namely $(\textrm{-}12,3),(\textrm{-}4,2),(2,\textrm{-}4),(3,\textrm{-}12)$. One solution $(0,0)$ is discarded because $x\neq 0$ and $y\neq 0$. Nine points on the graph are integral points.

Answer: $9$