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Solving a three-variable Diophantine |
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| Jun25-12, 10:31 AM | #1 |
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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? |
| Jun27-12, 05:44 AM | #2 |
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One nice solution: x=56, y=209, z=23409
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| 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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