
Re: Integer pairs in sum of reciprocals
Posted:
Jan 2, 2001 10:04 PM


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/

