The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Diophantine Equation

Date: 08/25/2007 at 17:46:32
From: Alex
Subject: an integer solution to (1-ab-a-b)/(1-ab+a+b)

Is there an integer solution to (1-ab-a-b)/(1-ab+a+b)?

I've been trying to do research on divisibility problems in my number 
theory books, but I'm still not even sure how to approach this type of 
problem.  I would appreciate it if you could give me a clue as to how 
to solve this kind of problem.

Date: 08/28/2007 at 20:10:47
From: Doctor Vogler
Subject: Re: an integer solution to (1-ab-a-b)/(1-ab+a+b)

Hi Alex,

Thanks for writing to Dr. Math.  I would write this as

  (1-ab-a-b)/(1-ab+a+b) = n


  (1-ab-a-b) = n(1-ab+a+b),

which is a Diophantine equation, and you can solve it using the method
described in

  Diophantine Equations with Three Variables 

More specifically, if you set

  x = a-1
  y = b-1
  z = n-1,

then your Diophantine equation is equivalent to

  xyz = 2x + 2y + 2z + 4.

Written this way, you can see that it is symmetric in x, y, and z
(indeed, the original equation is symmetric in a, b, and n), which
means that any integer solution (x, y, z) that you find can also be
permuted to find other integer solutions.  And if you permute the
solutions that you find, then you can limit your search to solutions
with a certain order.  For example, if you want to find all solutions
where (at least) one of the three variables is zero, then you might as
well look for the solutions where x = 0.  Then

  0 = 2y + 2z + 4


  z = -y - 2,

so you get the (infinitely many) solutions

  (0, k, -k - 2)

for any integer k, as well as the permutations

  (k, 0, -k - 2)


  (k, -k - 2, 0).

(You really should also list three more permutations, such as (-k, -2,
0, k), but this is just the second one again with a different choice
of k.)  Adding 1 to each number gives an (a, b, n) solution.

If none of the variables are zero, then you can divide by xyz and get

  1 = 2/(yz) + 2/(xz) + 2/(xy) + 4/(xyz).

Well, if x, y, and z are large, then the right side can't add up to 1.
For example, if all three have absolute value bigger than 2, then
that means that

  1 = 2/(yz) + 2/(xz) + 2/(xy) + 4/(xyz)
    = abs(2/(yz) + 2/(xz) + 2/(xy) + 4/(xyz))
    <= abs(2/(yz)) + abs(2/(xz)) + abs(2/(xy)) + abs(4/(xyz))
    <= 2/(3*3) + 2/(3*3) + 2/(3*3) + 4/(3*3*3)
    = 6/9 + 4/27
    < 1

which is impossible (1 < 1 is not true).  Therefore, at least one
variable must have absolute value 1 or 2.  Well, if x = 1 (or some
other value), then you can solve using the method of

  Solving a Diophantine Equation 

Alternately, you can just keep going with the previous idea.  For
example, if no variables are zero, and (at least) two of them have
absolute value bigger than 5, then you can't have a solution.  So now
you can try all (x, y) pairs with x and y between -5 and 5, and solve
for z in each one, that is, check if

  z = (2x + 2y + 4)/(xy - 2)

is an integer.  You'll find that apart from the
one-variable-equals-zero case (which I enumerated above), there are
only a small (finite) number of additional solutions.

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum 
Associated Topics:
College Number Theory

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.