Diophantine Equation with Three Unknowns

Date: 09/28/2009 at 22:48:52
From: Hue
Subject: Diophantine Equation with 3 unknowns

Find all triples of positive integers (x,y,z) such that

xyz + xy + yz + zx + x + y + z = 243

Here's what I've tried:

xyz + xy + yz + zx + x + y + z = 243 = 3^5

(xyz + xy + yz + zx + x + y + z)
--------------------------------- = 1
            243

 1       1        1        1        1        1        1        1
---- + ------ + ------ + ------ + ------ + ------ + ------ = ------
243     243z     243x     243y    243yz     243xz    243xy    xyz


 1       1        1        1        1        1        1        1
---- - ------ - ------ - ------ - ------ - ------ - ------ = ------
xyz     243z     243x     243y    243yz     243xz    243xy    243

If x, y, and z are each at least 2, then the L.H.S will be smaller
than the R.H.S which is less than 1/243.  So that means that at least
one of the variables must be smaller than 2, right?

If x = 1,

 1       1        1        1        1        1        1        1
---- - ------ - ------ - ------ - ------ - ------ - ------ = ------
 yz     243z     243      243y    243yz     243z     243y     243

 1       1        1        1        1        1        2        
---- - ------ - ------ - ------ - ------ - ------ = ------ 
 yz     243z     243y    243yz     243z     243y     243

 1          1         2        2        2             
---- (1 - ------) - ------ - ------ = ------
 yz        243       243y     243z     243     

From here I have no idea how to continue.

Thanks.

Date: 09/29/2009 at 02:21:05
From: Doctor Greenie
Subject: Re: Diophantine Equation with 3 unknowns

Hi, Hue --

Thanks for sending an interesting problem.  I didn't like the looks 
of the path you started down towards solving the problem, so I went 
a completely different direction.

  xyz + xy + xz + yz + x + y + z = 243
    x(yz + y + z) + (yz + y + z) = 243 - x
             (x + 1)(yz + y + z) = 243 - x
                      yz + y + z = (243 - x)/(x + 1)

The left-hand side must be an integer, so (x+1) must divide (243-x):

  (x+1) | (243 - x)
  (x+1) | (243 - x) + (x + 1)
  (x+1) | 244

This gives us only the following possible values for x+1:

  2, 4, 61, 122

which means x can only be

  1, 3, 60, or 121

We can then separate the problem into four cases with these values 
of x.  For each case, we encounter another problem that is solved by 
the same technique as above.

(1) x = 1...

    yz + y + z = (243 - 1)/(1 + 1) = 242/2 = 121
  y(z + 1) + z = 121
      y(z + 1) = 121 - z
             y = (121 - z)/(z + 1)

  (z+1) | (121 - z)
  (z+1) | (121 - z) + (z + 1)
  (z+1) | 122

  z + 1 = 2 or 61

  z = 1; y = (121-1)/(1+1) = 120/2 = 60

or

  z = 60; y = (121-60)/(60+1) = 61/61 = 1

So we have the solution (1,1,60) (and permutations of that solution, 
if we are looking for ordered triplets).

(2) x = 3...

    yz + y + z = 240/4 = 60
  y(z + 1) + z = 60
      y(z + 1) = 60 - z
             y = (60 - z)/(z + 1)

  (z+1) | (60 - z)
  (z+1) | (60 - z) + (z + 1)
  (z+1) | 61

  z + 1 = 61
      z = 60
      y = 0/61 = 0

The triple (0,3,60) gives us a solution to the equation, but it is 
not a solution in positive integers.  So we still have only the one 
solution.

I will let you finish the cases for x = 60 and x = 121.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 09/29/2009 at 05:05:56
From: Hue
Subject: Thank you (Diophantine Equation with 3 unknowns)

Hi Doctor Greenie,

Thanks for your guiding!!!

- Hue