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I am given a set $U = \{\alpha = a + b\sqrt2: N(\alpha) = a^2 - 2b^2 = 1\}$ where $a, b \in \Bbb{Z}$

I want to show that $3+2\sqrt2$ is the smallest element in this set that is greater than $1$.

To solve this, I have shown so far that only one of $\alpha$, $-\alpha$, $\alpha^{-1}$, and $-\alpha^{-1}$ can be greater than $1$. Also, $\alpha$ is greater than $1$ only when $a,b > 0$ .

How should I approach this? Thanks.

TheSimpliFire
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dmsj djsl
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  • In case you didn't know already, this is a question about the elements of norm $1$ (i.e. units) in the ring $\Bbb Z[\sqrt{2}]$, in which all the units are given by $\epsilon = \pm (1 + \sqrt{2})^n$ for $n \in \Bbb Z$. – Edward Evans Nov 29 '17 at 20:06

2 Answers2

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Solving Pell's equation $$ x^2-2y^2=1 $$ via continued fractions gives the fundamental solution $(x,y)=(3,2)$. The iterated solutions $(x_n,y_n)$ defined by $(x_1,y_1)=(x,y)$ and $(x_m,y_m)=(x_1+y_1\sqrt{2})^m$ give a monotonously increasing sequence $x_n+y_n\sqrt{2}$. So $3+2\sqrt{2}$ is the smallest one.

Possible duplicates:

Find all integer solutions to $x^2-2y^2=1$

Dietrich Burde
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A similar way w.r.t the first answer but maybe without any "fancy words" :)

You may simply plot $x^2 - 2y^2=1$ and look (and by look I really mean "look") at the plot and search for integer solutions near the origin. These are easily found:

There is $(1,0), (-1,0)$ and $(3,2)$ and all possible sign changes. All other integer solutions are obviously "far away". Then some simple arguments and case analysis (treating every "branch" of your plot separately) should do the trick, right?

Of course: If you have basic concepts of algebraic number theory at hand (or you are allowed to use them, since I don't know where you came across this problem) and something like Dirichlets Unit Theorem and so on then you should most likely make use of them and solve this problem say "algebraic number theoretically". My answer is only useful if you really have no idea of algebraic number theory.

M.U.
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