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!
Pell's equation
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elementary-number-theory
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1Do you know how to prove it has infinitely many solutions (so ignoring the k dividing y part)? – 2011-04-21
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0@Ross: I don't know how to start it! Thanks! And why do we need $k$? – 2011-04-21
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5As stated without background, this is a hard thing to do, so you should tell us what theorems you have seen in the course that you might be expected to use (I would suspect that you have proved it in the course without the $k$.) – 2011-04-21
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0If 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
2 Answers
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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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0Or am I missing something? – 2011-04-21
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2I 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-equation – 2011-04-21
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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.