Equation : x^2 + y^2 + z^2 = (x – y)(y – z)(z – x)

 
 
Show that the equation

x^2 \; + \; y^2 \; + \; z^2 \; = \; (x - y) \, (y - z) \, (z - x)

has infinitely many solutions in integers   x, \; y, \; z

 
 

Solution :

 
 

Let’s find solutions   (x, y, z)   which are in arithmetic progression.

y - x \; = \; z - y \; = \; d \; > \; 0
so that the equation reduces to the form

(y - d)^2 + y^2 + (y+d)^2
= 2 \, d^2 + 3 \, y^2

(y - x) \,(z - y) \,(z - x)
= \; (d) \,(d) \,(2 d)
= \; 2 \, d^3

2 \, d^2 + 3 \, y^2 \; = \; 2 \, d^3

Thus we get    3 \, y^2 \; = \; 2 \,(d - 1) \, d^2

2 \,(d - 1)   is 3 times a square.

This is satisfied if   d - 1 = 6 \, n^2   for some   n

Thus

d \; = \; 6 \, n^2 + 1    and    3 \, y^2 \; = \; d^2 \, 2 \,(6 \, n^2)

giving us

y^2 \; = \; 4 \, d^2 \, n^2

Thus we can take

y \; = \; 2 \,d \,n \; = \; 2 \,n \,(6 \,n^2 + 1)

From this we obtain

x \; = \; y - d \; = \; (2 \,n - 1) \,(6 \,n^2 + 1)
z \; = \; y + d \; = \; (2 \,n + 1) \,(6 \,n^2 + 1)

It is easily verified that

(x, y, z) \; = \; ((2 \,n - 1) \,(6 \,n^2 + 1), 2 \,n \,(6 \,n^2 + 1),(2 \,n + 1) \,(6 \,n^2 + 1))

is indeed a solution for a fixed   n   and this gives an infinite set of solutions as   n   varies over the set of integers.

 
 
 
 

 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

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

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