Diophantine equation: (x – y – z)^3 = 27 xyz

 
 
Are there any integral solutions   (x, \; y, \; z)   of the Diophantine equation

(x - y - z)^3 \; = \; 27 \, x \, y \, z

other than   (-n, \; n, \; n)   or such that   x \, y \, z \; = \; 0 ?

 
 

Solution:

Let

x \; = \; a^3
y \; = \; b^3
z \; = \; c^3

so that   a^3 - b^3 - c^3 \; = \; 3 \, a \, b \, c,   or equivalently

(a - b - c) \,((a + b)^2 + (a + c)^2 + (b - c)^2) \; = \; 0   ………..   (1)

since   (1)   it is equivalent to   2 \, (a^3 - 3 \, a \, b \, c - b^3 - c^3) \; = \; 0

Hence an infinite class of non-trivial solutions is given by

x \; = \; (b + c)^3
y \; = \; b^3
z \; = \; c^3

I don’t know whether or not there are any other solutions.

 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

2 Responses to Diophantine equation: (x – y – z)^3 = 27 xyz

  1. Paul says:

    Like these?, format {x, y, z}

    {-27,-8,-1}
    {-27,-1,-8}
    {-8,-27,1}
    {-8,1,-27}
    {-6,-48,6}
    {-6,6,-48}
    {-5,-40,5}
    {-5,5,-40}
    {-4,-32,4}
    {-4,4,-32}
    {-3,-24,3}
    {-3,3,-24}
    {-2,-16,2}
    {-2,2,-16}
    {-1,-27,8}
    {-1,-8,1}
    {-1,1,-8}
    {-1,8,-27}
    {1,-8,27}
    {1,-1,8}
    {1,8,-1}
    {1,27,-8}
    {2,-2,16}
    {2,16,-2}
    {3,-3,24}
    {3,24,-3}
    {4,-4,32}
    {4,32,-4}
    {5,-5,40}
    {5,40,-5}
    {6,-6,48}
    {6,48,-6}
    {8,-1,27}
    {8,27,-1}
    {27,1,8}
    {27,8,1}

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