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Let $d > 1$ be a squarefree integer. Prove that the equation $x^{2} - dy^{2} = 1$ has infinitely many solutions in $\mathbb{Z} \times \mathbb{Z}$.

What I have done: let $ \ \mathbb{K} = \mathbb{Q}[\sqrt{d}]$. If $ d \not\equiv 1$ mod $4$, then $O_{\mathbb{K}} = \mathbb{Z}[\sqrt{d}]$ and the statement follows by the Dirichlet Unit Theorem. What about the case $ d \equiv 1 $ mod $4$ ?

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  • $\begingroup$ Do you know that there is at least one non-trivial solution? $\endgroup$ – Thomas Andrews Aug 4 '13 at 15:39
  • $\begingroup$ yes : $x = 1$, $y = 0$ $\endgroup$ – WLOG Aug 4 '13 at 15:40
  • $\begingroup$ Yeah, I added "non-trivial." Sorry. $\endgroup$ – Thomas Andrews Aug 4 '13 at 15:41
  • $\begingroup$ By Dirichlet there are infinite units, their norms are $\pm 1 $ so in the case $ d \not\equiv 1$ mod $4$ we can always obtain a non-trivial solution $\endgroup$ – WLOG Aug 4 '13 at 15:45
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    $\begingroup$ Since $(x^2+dy^2)^2 - d(2xy)^2 = (x^2 - dy^2)^2$ If you find one non-trivial solution, you find infinitely many of them. $\endgroup$ – achille hui Aug 4 '13 at 16:12
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You can handle the case $d\equiv 1 \pmod{4}$ in essentially the same way. For any unit $u$, $u^6=a+b\sqrt{d}$ with $a,b\in\mathbb Z$ and $a^2-db^2=1$, then infinitude follows from Dirichlet's Unit Theorem.

If $u$ is a unit then $u^2=\frac{\alpha}{2}+\frac{\beta}{2}\sqrt{d}$ with $\alpha,\beta\in \mathbb Z$, $\alpha \equiv \beta \pmod{2}$ (by definition of $O_{\mathbb K}$) and $$ N(u^2)=\frac{\alpha^2}{4}-\frac{d\beta^2}{4}=1 $$ Then either:

Case 1 $\alpha\equiv\beta\equiv 0 \pmod{2}$. Then $\frac{\alpha}{2},\frac{\beta}{2}\in\mathbb Z$ and it follows from multiplication in $\mathbb K$ that $\exists a,b\in \mathbb Z$ with $u^6 =a+b\sqrt{d}$.

Case 2 $\alpha\equiv\beta\equiv 1 \pmod{2}$. Since $\alpha^2-d\beta^2=4$ and $\alpha^2\equiv\beta^2\equiv 1 \pmod{8}$ this case cannot arise unless $d\equiv 5\pmod{8}$. Then $$ u^6 = \frac{1}{8}\left(\alpha(\alpha^2+3d\beta^2)+\beta(3\alpha^2+d\beta^2)\sqrt{d}\right) \\ \alpha^2+3d\beta^2\equiv 3\alpha^2+d\beta^2\equiv 0 \pmod{8} \\ a=\frac{\alpha(\alpha^2+3d\beta^2)}{8},b=\frac{\beta(3\alpha^2+d\beta^2)}{8} $$ and $a,b\in\mathbb Z$.

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This is known as Pell's equation. As is explained in the Wikipedia article, one of the convergents of $\sqrt{d}$ (one of the rational approximations of the irrational number $\sqrt{d}$ found by truncating the continued fraction expansion) is a solution to the equation. From this solution all other solutions can be found - the way in which these solutions are generated shows that there are indeed infinitely many of them.

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    $\begingroup$ thank you, but i am interested in a solution using algebraic number theory but not continued fractions $\endgroup$ – WLOG Aug 4 '13 at 15:56
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If x solves $x^2-4 = dy^2$ then so does $x^2-2$

Put $(x^2-2)^2 = x^2 - 4x + 4$ , which is $d(xy)^2 + 4$

Dividing $x$, $y$ by $2$ will give this equation in units.

Now, if $x'$ and $x''$ are consecutive solutions, then $x x'' - x'$ is also a solution. Put

Since we have $2$ and $x$ gives $x^2-2$, it then follows that $x_{n+1} = x x_n - x_{n-1}$ os also a solution, by

$x_{n+1} + x_{n-1} = x x_n$

One shows that by this rather exotic arrangement, that if two values solve this equation, then the third one does also, and thus that this series is recursive over powers of $x$.

Suppose $a, b, c, d$ form a series such that $xb = a+c$. Then the measure $b^2 - ac$ is a constant, since $d = xc-b$.

Now, $c^2 - bd = c^2 - xcb + b^2 = (c-xb)c + b^2 = b^2-ac$

Thus, if eg $2, x, x-2, x^3-3x$ form such a series based on this solution

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    $\begingroup$ I think you are the only one who can understand what you wrote here. First of all, the equation is $x^2-1=dy^2$. Second, $(x^2-2)^2 = x^2 - 4x + 4$ looks like an elementary school mistake. Third... Fourth... $\endgroup$ – user26857 Aug 8 '13 at 15:33

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