Solution to Problem #24

Andrey Mamontov of Reedsburg High School in Reedsburg WI found this solution:

1/x+1/y=1/2000

(y+x)/(xy)=1/2000

2000(x+y)=xy

2000x+2000y-xy=0

xy-2000x-2000y+4*10^6=4*10^6

x(y-2000)-2000(y-2000)=4*10^6

(x-2000)(y-2000)=4*10^6

Each factorization ab=4*10^6 (where a and b are
integers) gives us a solution. We will say that
x-2000=a and y-2000=b. There are many different ways
for factorization. In order to find all possible
factorization, we split 4*10^6 into prime factors, for
example by the Euclid Algorithm, 4*10^6=(2^8)*(5^6).
The solution would be all possible combinations of
such numbers. In fact, because the number of prime
factors is finite, the number of possible combination
is finite, and thus the number of solutions is finite
as well.

Following Andrey's recipe we get these solutions with x less than or equal to y (the rest can be obtained by symmetry):

 x             y

2001        4002000
2002        2002000
2004        1002000
2005         802000
2008         502000
2010         402000
2016         252000
2020         202000
2025         162000
2032         127000
2040         102000
2050          82000
2064          64500
2080          52000
2100          42000
2125          34000
2128          33250
2160          27000
2200          22000
2250          18000
2256          17625
2320          14500
2400          12000
2500          10000
2625           8400
2640           8250
2800           7000
3000           6000
3250           5200
3280           5125
3600           4500
4000           4000

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