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Topic: Integer pairs in sum of reciprocals
Replies: 39   Last Post: Jan 22, 2001 6:02 PM

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 Dik T. Winter Posts: 7,899 Registered: 12/6/04
Re: Integer pairs in sum of reciprocals
Posted: Jan 2, 2001 10:04 PM
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In article <3A52637B.FA726CF@hot.rr.com> "Ross A. Finlayson" <rfinlayson@hot.rr.com> writes:
> Here is a regular form:
> 1/3 x + 1/6 x = 1/2 x
> Where here the a, b and c are 3x, 6x, and 2x, for any positve integer x.
> For any positive integer x, x=x, thus there are infinite solutions.
> There are probably infinite regular forms.

Yup. Not only probably but provable. Take p a prime, we have:
1/(p+1) + 1/p(p+1) = 1/p
and when you multiply all three demoninators by the same integer does not make
a difference. So there are inifinitely arithmetically unrelated forms.

However, in your first response you stated:
> Set a to 2001 or greater and b to a large value. The result is very close
> to 1/2001, where 2001 is called c,
> 1/a + 1/b = 1/c, or
> a ^-1 + b ^-1 = c^-1
> There are probably infinite solutions.

As David Eppstein has shown in this very same thread, for each fixed c there
are only finitely many solution, and the number of solutions is also shown
(and that number is never 0).
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

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