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Why is it that for the Diophantine equation
[eqn] \displaystyle x^2 - Dy^2 = 1 [/eqn]
(Pell's equation) with fundamental solution math] (x_1, y_1) [/math] will have all other solutions in the form:
[eqn] \displaystyle x_{k}+y_{k}{\sqrt {n}}=(x_{1}+y_{1}{\sqrt {n}})^{k} [/eqn]
?. Anything I can find online claims it as general knowledge and skips over the reason it is valid.
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>>8582528
What's [math] D [/math] supposed to be here? An operator or a constant?
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The Wolfram Alpha article on it
http://mathworld.wolfram.com/PellEquation.html
has "slightly" more info on it. I just don't see how they go from (32) to (33) and (34),
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>>8582548
It's a natural number
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theres a proof in An Introduction to the Theory of Numbers by niven/zuckerman/montgomery
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>>8582549
Agree it's confusing. I suggest just starting at 33. Imagine multiplying out the right hand side. You will get an expression in the form of the left hand side, giving you x and y. Do the same for 34.
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>>8582986
Suppose p^2 - D q^2 = 1.
Then expand out: (p +sqrt(D) q)^n =
(horrible expression) + (another horrible expression)*sqrt(D),
where the horrible expressions contain no sqrt(D)'s. Let
x=horrible expression
y=another horrible expression, so
(x+sqrt(D) y) = (p +sqrt(D) q)^n
Expand out (p-sqrt(D)q)^n, see that
(x+-sqrt(D) y) = (p -sqrt(D) q)^n
[this works becse (-sqrt(D))^2 = (sqrt(D))^2=D
Multiply these two expressions to get
(x^2-Dy^2) = (p^2-sqrt(D) q^2)^n =1^n = 1.
So (x,y) is another solution.
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