If know that $x=3$, $y=2$ is a solution of $$x^2-2y^2=1,$$ then apparently all other solutions can be calculated as $$x_k+y_k\sqrt{2}=(x+y\sqrt{2})^k,$$ which I have trouble understanding. I've been told that it can be seen thorugh the continued fraction of $\sqrt{2}$, but I don't see it. Could someone explain to me what exactly the idea is and how I prove this?
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Not sure how much I want to explain, given how many times I have answered this question, but all solutions $(x,y)$ with non-negative entries can be found by beginning with $(1,0)$ and then finding the next pair with $$ (3 x + 4 y, 2 x + 3 y). $$ $$$$ $$ (1 ,0 ), $$ $$ ( 3,2 ), $$ $$ (17 ,12 ), $$ $$ ( 99,70 ), $$ $$ (577 ,408 ), $$ $$ ( 3363,2378 ), $$ $$ ( 19601,13860 ), $$ $$ (114243 ,80782 ), \ldots $$
It is also true that $(x_n, y_n)$ obey $$ x_{n+2} = 6 x_{n+1} - x_n, $$ $$ y_{n+2} = 6 y_{n+1} - y_n. $$
Will Jagy
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Thank you for your answer! Could you briefly explain where the continued fraction of $\sqrt{2}$ comes into play? – Zolf69 Sep 14 '14 at 21:28