Triples of Numbers

Date: 7/23/96 at 10:24:53
From: Anonymous
Subject: Triples of Numbers

Find two triples of positive integers such that their sum equals their 
product. I found one right away (1,2,3). I insist that there is no way 
there can be another one. Is there? And if there is not, can you prove 
it?

Date: 7/26/96 at 15:41:36
From: Doctor Ceeks
Subject: Re: your response

Hi,

There are more: (0,A,-A) works...both product and sum equal 0.
Are you restricting yourself to integers?  If not, then there
are many, many, many solutions.

If you are considering only positive integers, then you've found the 
only one.  To see this: you want a,b,c such that a+b+c=abc. 
Rearranging, get c=(a+b)/(ab-1).  For c to be an integer, we must have 
ab-1<= a+b.  If both a and b are bigger than 2, then 
ab > max(a,b)*2 >= a+b, so we would need equality to occur in the 
second inequality...that is a=b, but there is no solution with a=b 
(check that 2a/(a^2-1) is never an integer).  So one of a or b is 
equal to 1 or 2.

Case 1: a = 1.  c =(b+1)/(b-1) is an integer only when b=2 or 3, in 
which cases c = 3 or 2, respectively.

Case 2: a = 2.  c =(b+2)/(2b-1) is an integer only when b=1 or 3, in
which cases, c=3 or 1, respectively. (There are some small things to 
check above, but this outline is correct.)

-Doctor Ceeks,  The Math Forum
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