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Diophantine Equation with Three Variables

Date: 02/01/2006 at 02:25:21
From: Pranjal
Subject: Integer Solution of 3 variables

I want to find all positive integer solutions for a + b + c + ab + bc
+ ca = abc + 1.

I guessed a lot and found a solution of 3,3,7 but that isn't logical
and I don't know other solutions.  I can't find any other relationship
for the variables.  How can I solve 1 equation for 3 variables?

Date: 02/01/2006 at 19:00:31
From: Doctor Vogler
Subject: Re: Integer Solution of 3 variables

Hi Pranjal,

Thanks for writing to Dr. Math.  This is called a "Diophantine
equation", and I love solving them.  Let me give you some ideas on how
to approach such a problem.

Sometimes a search is exactly the correct thing to do, but the trouble
is that you can't try every (positive) integer, so you need a little
more mathematical analysis to prove that your search has found *all*
of the solutions.  Sometimes this is very difficult to do, but it is
possible for many kinds of Diophantine equations.

Your equation has the special property that there is one large term
(abc) that is a multiple of every other term.  When this happens,
there is a technique you can use to give a bound on the solutions. 
The idea is to divide by that term:

  1 = 1/(bc) + 1/(ac) + 1/(ab) + 1/c + 1/a + 1/b - 1/(abc).

Now consider:  If a, b, and c are all large numbers, then each of the
fractions on the right will be small.  But if they are too small, then
they can't sum to 1.  In particular, if a, b, and c are each at least
4, then the right side will be smaller than

  1/16 + 1/16 + 1/16 + 1/4 + 1/4 + 1/4

which is less than 1.  So that means that at least one of the
variables must be smaller than 4.  (Also notice that a, b, and c being
positive is not all that important here, since if a, b, and c each
have absolute value at least 4, then the right side of the equation
has absolute value at most

  1/16 + 1/16 + 1/16 + 1/4 + 1/4 + 1/4 + 1/64 < 1.)

One way to use this fact is to try the four cases a=0, a=1, a=2, and
a=3, and see what equation is left.  (You have three more cases if you
allow negative solutions.)  Then you can do the same kind of analysis
to get upper bounds for b and c.

But we can use another idea to do even better.  Notice that if your
equation were

  1 + ab + bc + ca = abc + a + b + c,

then we could factor this as

  (a - 1)(b - 1)(c - 1) = 0.

And this is easy to solve; it only needs one of the three variables to
equal 1.  Well, your equation doesn't factor so nicely, but that
product uses up the degree-2 and the degree-3 terms of your equation.
So if we make the change of variables

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

then your equation becomes

  xyz - 2x - 2y - 2z - 4 = 0


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

The important thing here is that every denominator has at least two
variables.  So you could say that if each has absolute value at least
3, then the right side will have absolute value at most

  2/9 + 2/9 + 2/9 + 4/27 < 1,

but it is even more useful to say that as long as *two* of the three
variables has absolute value at least 6, the right side will have
absolute value at most

  2/6 + 2/6 + 2/36 + 4/36 < 1.

So we can conclude that of x, y, and z, not only one but two variables
must have absolute value less than 6.  So you can try all pairs of
values (x, y) with x and y of absolute value less than 6, and see
which ones give an integer for

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

Then you can change the solutions back to

  a = x + 1
  b = y + 1
  c = z + 1.

I find a few more solutions besides the one you mentioned, such as

  (2, 4, 13).

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

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