# 2005

phgao
Posers and Puzzles 04 May '05 06:59
1. 04 May '05 06:59
Find all pairs of positive integers x and y that safisfy the equation:

1/x - 1/y = 1/2005
2. PBE6
Bananarama
04 May '05 15:57
OK, here's what I got:

x = 1600; y = 8020
x = 1980; y = 158796
x = 2000; y = 802000

x = 2004; y = 0 is another pair, but 0 is not a positive integer.

Sadly, I solved this by trial and error. :'( If anyone knows how to do this analytically, I'd love to find out how.
3. Acolyte
04 May '05 17:34
Originally posted by PBE6
OK, here's what I got:

x = 1600; y = 8020
x = 1980; y = 158796
x = 2000; y = 802000

x = 2004; y = 0 is another pair, but 0 is not a positive integer.

Sadly, I solved this by trial and error. :'( If anyone knows how to do this analytically, I'd love to find out how.
Anayltically? I don't think there's a magic method of doing this, but it's easy to show rigorously that you've found all the solutions: all you need to do is remark that for all solutions in positive integers,
1/x < 1/2005, so x < 2005,
and find y for each x in turn, confirming that the solutions you listed below are the only ones in positive integers.

1/2004 -1/0 is -infinity though, so that's not a solution at all.
4. PBE6
Bananarama
04 May '05 17:52
Originally posted by Acolyte
Anayltically? I don't think there's a magic method of doing this, but it's easy to show rigorously that you've found all the solutions: all you need to do is remark that for all solutions in positive integers,
1/x < 1/2005, so x < 2005,
and find y for each x in turn, confirming that the solutions you listed below are the only ones in positive integers.

1/2004 -1/0 is -infinity though, so that's not a solution at all.
Oops, my mistake (with the 0).
5. 04 May '05 19:172 edits
Originally posted by PBE6
OK, here's what I got:

x = 1600; y = 8020
x = 1980; y = 158796
x = 2000; y = 802000

x = 2004; y = 0 is another pair, but 0 is not a positive integer.

Sadly, I solved this by trial and error. :'( If anyone knows how to do this analytically, I'd love to find out how.
I think your 1600 should be 1604.

x must be < 2005; so we need to find those values of x such that y = 2005x/(2005 - x) is also an integer.

Here's what I get for (x,y), and I think this is all such pairs:

(1604, 8020)
(1980, 158796)
(2000, 802000)
(2004, 4018020)

EDIT: I don't see any magical analytical solution either, although the problem can be whittled down quite a bit.
6. PBE6
Bananarama
04 May '05 20:04
Originally posted by davegage
I think your 1600 should be 1604.

x must be < 2005; so we need to find those values of x such that y = 2005x/(2005 - x) is also an integer.

Here's what I get for (x,y), and I think this is all such pairs:

(1604, 8020)
(1980, 158796)
(2000, 802000)
(2004, 4018020)

EDIT: I don't see any magical analytical solution either, although the problem can be whittled down quite a bit.
Oops again. I think my blood sugar is low. I need a cookie.

7. 04 May '05 22:29
Originally posted by PBE6
Oops again. I think my blood sugar is low. I need a cookie.

Let a,b be relatively prime integers , and x=ka and y=kb, b>a

then k=c(b-a) where abc=2005 will give all the solutions:

x=(b-a)ca, y=(b-a)cb.

When c=1 there are two sets of factors to use for ab;
when c=5 or 401 there is just one set of factors for ab.
c cannot be 2005 as x=y not possible.

There are just these four since the only factors of 2005
are 1, 5, 401, and 2005.

[ 2004(1) , 2004(2005) ] , [ 396(5) , 396(401) ]

[400(5)(1) , 400(5)(401) ] , [ 4(401)(1) , 4(401)(5) ]