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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?


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I explained how you can get nontrivial solutions in… – Zarrax Jan 26 '11 at 15:41
@Zarrax: First of all, Thanks. I read that thread, however, in my case, it asked for proving integers solution. Plus 13 is irrational. – Chan Jan 26 '11 at 15:47
ok, I misread what you needed. (By the way 13 is not irrational) – Zarrax Jan 26 '11 at 15:51
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 $\sqrt{13}$. – Chris Eagle Jan 26 '11 at 15:59
@Zarrax: Thanks, as for 13, I mean sqrt(13). Sorry! – Chan Jan 26 '11 at 16:10
up vote 5 down vote accepted

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,


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 $\displaystyle r = 5$.


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. – Chan Feb 1 '11 at 4:09

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