What Are the Solutions to the Diophantine Equation x^2 = 2y^2 = 3z^2?

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Homework Statement


x^2 = 2y^2 = 3z^2
Find all the solutions.


Homework Equations


There is a method once I substitute the above formula into the form ap^2 + q^2 = z^2, to get all the solutions.


The Attempt at a Solution


I'm having a little trouble with the substitution.
I was thinking, at first that one could argue that x and y are both congruent to 0, mod 3. Then x-y = 3p and x+ 2y = 3q; one could solve for x, and y, and then backsubstitute, when I get the formulas for x and y in terms of integers, they don't check in the first equation
 
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lei123 said:

Homework Statement


x^2 = 2y^2 = 3z^2
Are you sure this is correct? Is it supposed to be ##x^2 + 2y^2 =3z^2## or something similar?
##x^2 = 2y^2 = 3z^2## does not seem correct, maybe it's a typo.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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