Solving a three-variable Diophantine

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SUMMARY

The discussion focuses on solving the three-variable Diophantine equation $$(4x^2+1)(4y^2+1) = (4z^2+1)$$ for positive, nonzero integers x, y, and z. A notable solution provided is (x=56, y=209, z=23409). The user suggests a method where if x is a natural number, then setting y as \(4x^2\) leads to the formula \(z = x(2y + 1)^2\), yielding several solutions such as (1,4,9), (2,16,66), and (3,36,219).

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SeventhSigma
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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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One nice solution: x=56, y=209, z=23409
 
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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