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Solve $$ 3x^2 - 2y^2 =1 $$ in $ \mathbb{Z}$.

How can we do it?

( All of answers gave me a great help. Thanks a lot kind stackexchangers.)

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    Have you tried anything at all? For example, it is *obvious* that $\,x\,$ must be odd...or isn't it?2012-08-12
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    Yes $$x$$ must be odd......2012-08-12
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    How about relating it to Pell's equation? I found that any pair of integers $(x_n, y_n)$ satisfying $$3x_n + y_n\sqrt{6} = (5+2\sqrt{6})^n (3+\sqrt{6})$$ is a solution to the given equation, and I suspect that (up to change of sign) these are the solutions.2012-08-12
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    Relating Pell's equation 을 어떻게 하는지 그 과정을 잘 모르겠습니다.. (It is the point that I am very wondering...) Would you give me a kind derivation, please?.. Thank you in advance.2012-08-12
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    I'm afraid I am bad at number theory. I know how to construct a family of solutions via fundamental solutions, but I do not know how to prove that these exhaust all the possibility. I believe someone else will give you a nice description in this way, or in a better way.2012-08-12
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    @sos440: For me it is slightly smoother to multiply by $2$, and rewrite as $(2y)^2-6x^2=-2$, then $w^2-6x^2=-2$. Because $|-2|\lt \sqrt{6}$, standard theory says that any solution is obtained from a convergent $\frac{r}{s}$ of the continued fraction expansion of $\sqrt{6}$. From this we can show that there is only one "class" for $-2$. Note that $w=2$, $y=1$ is a solution. So all solutions are of shape $(a,b)$ where $a+b\sqrt{6}=(2+\sqrt{6})(5+2\sqrt{6})^n$.2012-08-12

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