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Please help me with the following question. Thank you! We know that $D$ is a positive integer, not a square. We let $k$ be any positive integer. We need to prove that the equation $x^2 - D y^2 = 1$ has infinitely many solutions with $k$ dividing $y$. Any help is appreciated! Thanks!

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    If you look at http://en.wikipedia.org/wiki/Pell%27s_equation, the section entitled "Additional solutions from the fundamental solution" you will see how to make a chain of solutions via a set of recurrence relations. Can you justify that lots of them have $k|y_n$? Maybe work modulo $k$?2011-04-21

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Set D' = k^2 D and consider the equation

x^2 - D' y^2 = 1

which is known to have infinite solutions...

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    I was about to say that the theory of the Pell equation assumes $D$ squarefree, but it is an inessential restriction, as discussed in http://math.stackexchange.com/questions/19177/non-squarefree-version-of-pells-equation2011-04-21
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Define two sequences of polynomials $f_0 = 2, f_1 = X, f_n = X f_{n-1} - f_{n-2},$ $g_0 = 0, g_1 = 1, g_n = X g_{n-1} - g_{n-2}.$ If $(X,Y)$ satisfies $X^2 - D Y^2 = 4$, then so does $(f_n(X), Y \cdot g_n(X))$ for all possitive integers $n$. This applies to $x^2 - D y^2 = 1$ by letting $X = 2 x$ and $Y = 2 y$. Assume $k$ divides $y$. Then $k$ divides $y$ for each $n > 0$. This supplements Aryabhata's excellent answer by showing one way we can get infinitely many solutions from one.