Equation: (x^2 + 1)(y^2 + 1) = (z^2 + 1)

 
 
Does there exist an infinite number of solutions to the equation:

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

where each of   x, \; y   and   z   is a positive integer > 1

 
 
Here are two parametric solutions:

x \; = \; n
y \; = \; n \; + \; 1
z \; = \; n^2 \; + \; n \; + \; 1

(n^2 + 1) \, ((n+1)^2 + 1) \; = \; (n^2 + n + 1)^2 \; + \; 1

x \; = \; n
y \; = \; 2 \, n^2
z \; = \; 2 \, n^3 \; + \; n

(n^2 + 1) \,((2 \, n^2)^2 + 1) \; = \; (2 \, n^3 + n)^2 \; + \; 1

 

There may be other parametric solutions.
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

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

3 Responses to Equation: (x^2 + 1)(y^2 + 1) = (z^2 + 1)

  1. paul says:

    I’m going with yes.

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