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

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About mvtrinh

Retired high school math teacher.
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