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My problem is proving $x^2 - 13y^2 = 1$ has integers solutions. I can find easily see that (+-1, 0) are trivial solution. My question is: is it sufficient to complete the proof? How can I approach this problem?

Thanks,
Chan

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    I explained how you can get nontrivial solutions in http://math.stackexchange.com/questions/13430/does-there-exist-x-y-in-mathbbq-mathbbz-such-that-x22y2-12011-01-26
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    @Zarrax: First of all, Thanks. I read that thread, however, in my case, it asked for proving integers solution. Plus 13 is irrational.2011-01-26
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    ok, I misread what you needed. (By the way 13 is not irrational)2011-01-26
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    You have already shown $x^2-13y^2=1$ has integer solutions. It has nontrivial solutions by the Dirichlet unit theorem. One way to find them is by looking at the convergents of the continued fraction of $\sqrt{13}$.2011-01-26
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    @Zarrax: Thanks, as for 13, I mean sqrt(13). Sorry!2011-01-26
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    @Chris Eagle: Thanks. So it was good enough to complete the proof right? But when I think over it, it's more like finding integers solution instead of proving it.2011-01-26
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    You can use http://www.alpertron.com.ar/QUAD.HTM to find a recurrence which gives all solutions2011-01-26
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    @Chan: To prove that an equation has (nontrivial) integer solutions, it suffices to produce at least one (nontrivial) integer solution; it does not matter how one finds that solution: it can be obtained through a general procedure valid for all kinds of similar problems, or it can be obtained through fasting, meditation, and divine-inspiration-through-your-deity-of-choice. For a proof of existence, it is completely irrelevant *how* you find the solutions, so long as they *are* solutions. (It's possible to prove existence without producing a solution, but producing a solution is *enough*)2011-01-26

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Presuming you meant an infinite number of solutions, Pell's equation can be solved using the theory of continued fractions and the theory shows that $\displaystyle x^2 - dy^2 = 1$ always has an infinite number of solutions for irrational $\displaystyle \sqrt{d}$ ($\displaystyle d$ a positive integer).

The following are well known, for $\displaystyle d$ not a perfect square,

1)

The continued fraction expansion of $\displaystyle \sqrt{d}$ is periodic, say with period $\displaystyle r$.

For instance in your case

$$\sqrt{13} = (3, \overline{1, 1, 1, 1,6})$$

and $\displaystyle r = 5$.

2)

If $\displaystyle \frac{h_m}{k_m}$ is the $m^{th}$ convergent of $\displaystyle \sqrt{d}$, which is of period $\displaystyle r$, then we have that

$$(h_{nr-1})^2 - d \ (k_{nr-1})^2 = (-1)^{nr}$$

Thus in case of $\displaystyle \sqrt{13}$, we have that for any even $\displaystyle n = 2t$

$$(h_{10t-1})^2 - d \ (k_{10t-1})^2 = 1$$

which for instance, gives us the following as a solution for $t=1$:

$$649^2 - 13 \times 180^2 = 1$$

Note one can go further and prove that the continued fractions help us generate all the solutions of the Pell's equation.

I would recommend you read the excellent chapter on Continued Fractions in the book: An Introduction to the Theory of Numbers by Niven and Zuckerman.

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    Many thanks for your clear explanation.2011-02-01