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### 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

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
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra

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