How to find the fundamental unit in $\mathbb Z[\sqrt{19}]$? I have no idea how it may look like.
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Look up Pell's equation and continued fractions. – Ethan Bolker Dec 24 '17 at 11:26
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Why don't you try calculating $$\sqrt{\frac{n^2 \pm 1}{19}}$$ starting at $n = -169$ and incrementing $n$ to $-168, -167, -166$, etc., until you find an integer? Mwahahaha! – The Short One Dec 26 '17 at 21:55
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Observe that $\alpha = 12 + \sqrt{19}$ has norm $5^3$; now look for elements of norm $\pm 5$. You will find $\beta = 9 + 2\sqrt{19}$ is such an element. Thus $$ \varepsilon = \alpha/\beta^3 = 170 - 39\sqrt{19}$$ is a nontrivial unit.
It remains to check that $\varepsilon$ is not a power of another unit. It cannot be a square since $39$ is odd. If you set $\varepsilon = \gamma^n$ for $\gamma = c + d \sqrt{19}$, then $d \mid 39$, and you can quickly eliminate all possible cases.
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You might like Prof. Lubin's method, which he says he got after teaching a class and summarizing what he had done on the blackboard. I think it is, likely, very close to what Wallis, Brouncker, Fermat would have done. I wrote it up as a fairly simple C++ program. Almost forgot https://math.stackexchange.com/questions/2215918/continued-fraction-of-sqrt67-4/2216011#2216011 I see, I typeset two of these at the same question – Will Jagy Dec 24 '17 at 20:12
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To find the fundamental unit of $\mathbf Z\bigl[\sqrt{19}\mkern1mu\bigr]$, you have to find the smallest non-trivial solution in positive integers of the Pell-Fermat equation: $$x^2-19y^2=1.$$ The tool for this is to find the expansion of $\sqrt{19}$ as a continued fraction. The expansion of a quadratic number is periodic, and if it has period $n$, it is denoted as $$\bigl[a_0;\overline{a_1a_2\dots a_n}\bigr]$$ Lagrange proved the fundamental unit is the convergent of order $n-1$.
Using Pari-GP, we find that $$\sqrt{19}=\bigl[4;\overline{213128}\mkern 1mu\bigr],$$ hence the $x$ and $y$ of the fundamental unit are the numerator and denominator of the convergent of order $5$: $$4+\frac{1}{2+\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{1+\cfrac{1}{2}}}}}=\frac{170}{39}$$ Thus the fundamental unit (in the first quadrant) is $\;\color{red}{170+39\sqrt{19}}$.
Bernard
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