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Let $\mathbb{Q}(\sqrt{d})$ be a real quadratic field. If $d \equiv 2, 3$ modulo 4, then the ring of integers is $\mathbb{Z}[\sqrt{d}]$ and, if you want to compute the group of units, you need to find the minimal solution to Pell's equation $x^2-dy^2=\pm 1$. How to do this in terms of continued fractions is explained in all standard books.

However, there is also the case $d \equiv 1$ modulo 4. Then the ring of integers is $\mathbb{Z}[\frac{1+\sqrt{d}}{2}]$ and the equation becomes $x^2-dy^2=\pm 4$, again for $x, y$ integers. I have not found any reference about how to find a solution to this one. Of course, if you have a solution $(x_0, y_0)$ to the first one, then $(2x_0, 2y_0)$ is a solution of the second one, but it is never going to be the minimal one (I think) since it lies in $\mathbb{Z}[\sqrt{d}]$. For instance, if $d=5$, the minimal solution is $\frac{1+\sqrt{5}}{2}$.

So how does the method work in this case?

Martin Sleziak
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    Should be explained in H. Cohen "Number Theory: Volume I: Tools and Diophantine Equations", Section 6.3 – Max Alekseyev Oct 16 '16 at 15:24
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    The fundamental solution of $x^2-dy^2=1$ is either the fundamental unit or its cube. The fundamental unit can be found using the continued fraction expansion of $(1 + \sqrt{d})/2$ in much the same way as for $x^2-dy^2=1$. Yes, this is all "well known". – Noam D. Elkies Oct 16 '16 at 15:29
  • Thanks for the reference to Cohen, it is indeed well explained there –  Oct 16 '16 at 15:54

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