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
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
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.