
Re: Integer pairs in sum of reciprocals
Posted:
Jan 2, 2001 1:08 PM


On Tue, 2 Jan 2001, Ross A. Finlayson wrote:
> Jan Kristian Haugland wrote: > > > "Ross A. Finlayson" wrote: > > > > > saxon970@yahoo.com wrote: > > > > > > > Hello. I have some questions about how to approach the problem below: > > > > > > > > How many pairs of positive integers a and b are there such that a < b > > > > and > > > > 1/a + 1/b = 1/2001 ? > > > > > > > > End of problem. > > > > > > > > Is trial and error and making a manual list the way to go? > > > > > > > > Thanks. > > > > > > > > Sent via Deja.com > > > > http://www.deja.com/ > > > > > > 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. > > > > What does this mean? (1) There are infinitely > > many triplets (a, b, c) with 1/a + 1/b = 1/c, > > or (2) there are infinitely many pairs (a, b) > > with 1/a + 1/b = 1/2001? If a, b and c are > > restricted to the positive integers then (1) > > is trivially true and (2) is trivially false. > > > > > Ross > > > > > >  > > > Ross Andrew Finlayson > > > Finlayson Consulting > > > Ross at TikiLounge: http://www.tikilounge.com/~raf/ > > > "The best mathematician in the world is Maplev in Ontario."  Pertti L. > > > >  > > > > Jan Kristian Haugland > > http://home.hia.no/~jkhaug00 > > If you assign c a sufficiently large value, then for some value a slightly > greater than c and b greater than a then there is a solution.
Do you have a proof for this statement?
> If there is one there is probably infinite.
There is one positive integer less than 2. Does that mean there are an infinite number of them?
Steve L

