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Solving a three-variable Diophantine

 
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Jun25-12, 10:31 AM   #1
 

Solving a three-variable Diophantine


I have the following equation

$$(4x^2+1)(4y^2+1) = (4z^2+1)$$

For positive, nonzero integers x and y (and thus z). I am having difficulty figuring out a good method/algorithm for calculating solutions to this equation. Any thoughts?
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Jun27-12, 05:44 AM   #2
 
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One nice solution: x=56, y=209, z=23409
Jun27-12, 10:38 AM   #3
 
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A 'good deal' of the solutions are caught by:

let x be a natural number and y = [itex]4 x^{2}[/itex],

then we have z = [itex]x (2 y +1)^{2}[/itex]

for example (x,y,z) = (1,4,9), (2,16,66), (3,36,219), ..., (17,1156,39321)
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