
Re: Integer pairs in sum of reciprocals
Posted:
Jan 11, 2001 12:08 PM


In article <3A5139E4.94731FAA@hot.rr.com>, "Ross A. Finlayson" <rfinlayson@hot.rr.com> wrote: > > saxon970@yahoo.com wrote: > > > > How many pairs of positive integers a and b are there such that > > a < b and 1/a + 1/b = 1/2001 ? > > 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.
saxon's message has disappeared from Deja, but as several people mentioned (including Nathan until he changed his mind!) there are 13 solutions:
Writing 1/a + 1/b = 1/2001 as u.v = 2001^2 = 3^2.23^2.29^2, with u, v = a  2001, b  2001, there are 3^3 = 27 integer pairs u, v.
(Each 3 here is for the 3 possible powers 0, 1, 2 of each prime independently that can divide say u. More generally, if the number was p^m.q^n.. with p, q,.. prime then the number of pairs would be (m+1).(n+1)..).
If u, v < 0 then for 0 < a, b we require u, v < 2001 which is impossible. Thus 0 < u, v.
There are thus 13 such pairs with 0 < u < v, i.e. half of the 26 pairs with u not equal v.
Cheers
John_Ramsden@acer.de
Sent via Deja.com http://www.deja.com/

