I want to prove this statement : $a+b \sqrt{2}=(1+ \sqrt{2})^n$ when $a^2-2b^2=+1 or -1$ Where a,b,and n are positive integers . Usually people consider it true saying $1+ \sqrt{2}$ is fundamental unit of $\mathbb{Z} [ \sqrt{2}]$ , but how can I proof this more rigorously ??
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Using mathjax will help your questions get better responses https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Lee Mosher Dec 25 '17 at 15:02
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"$1+\sqrt 2$ is fundamental unit of $\mathbb Z[\sqrt 2]$" is a rigorous proof(though a bit circular). – Cave Johnson Dec 25 '17 at 15:04
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I suggest these notes from Dusan Djukic: www-bcf.usc.edu/~lototsky/PiMuEp/Pell-IMO.pdf – Jack D'Aurizio Dec 25 '17 at 15:20
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1You left out an indication of what kind of integers $a, b$, and $n$ are! If they are all just integers then the claim is false: $a = -3, b = -2$ fit but $-3-2\sqrt{2} \not= (1+\sqrt{2})^n$ for any integer $n$ since the left side is negative and the right side is positive. If you are trying to exactly classify the integer solutions $a, b$ to $a^2-2b^2 = 1$ then the correct formula uses the coefficients of $\pm(1+\sqrt{2})^{2n}$, or equivalently $\pm(3+2\sqrt{2})^n$, for all $n \in \mathbf Z$. – KCd Dec 25 '17 at 15:26
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Oops I missed that i'll edit it thanks – Rz_CHIFFRE Dec 25 '17 at 15:28
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The edited task is still incomplete: if $a$ and $b$ are positive integers satisfying $a^2-2b^2=1$ then $a+b\sqrt{2} = (1+\sqrt{2})^n$ for some positive integer $n$, but not conversely: try $n= 1$, or really any odd positive integer $n$. Either consider the equations $a^2-2b^2=\pm 1$ together (that corresponds to all positive integers $n$) or restrict to even positive integers $n$ (equivalently, look at positive integer powers of $3+2\sqrt{2}$). – KCd Dec 25 '17 at 15:34
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Since all elements $(1+\sqrt{2})^n$ are invertible in $\mathbb{Z}[\sqrt{2}]$, because of $-(1+\sqrt{2})(1-\sqrt{2})=1$, the norm of such elements is equal to $\pm 1$. But the norm is $\pm 1=N(a+b\sqrt{2})=a^2-2b^2$.
(Rigorous) Reference: The units of $\mathbb Z[\sqrt{2}]$
Dietrich Burde
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To prove this is equivalent to proving it's the fundamental unit. Here's a hint for a direct proof, based on understanding how you might show some element is the generator of a cyclic group.
The powers of $\mu = 1 + \sqrt{2}$ form an increasing unbounded sequence of real numbers. Let $\nu$ be a positive unit. Find the largest power of $\mu$ that's less than or equal to $\mu$. Divide $\nu$ by that power to produce a small unit.
Ethan Bolker
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Would that work if I showed (Z[sqrt(2)],.) Is a monogenous group ?? – Rz_CHIFFRE Dec 25 '17 at 15:14
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I've never seen the term "monogenous group". You need to show there's a smallest positive unit, and that $\mu$ is it. The norm map helps. – Ethan Bolker Dec 25 '17 at 15:26
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Define $a_n$ and $b_n$ by \begin{eqnarray*} a_n+b_n \sqrt{2} =(1+\sqrt{2})^n. \end{eqnarray*} Their recurrence relation is easily derived by \begin{eqnarray*} a_{n+1}+b_{n+1} \sqrt{2} =(a_n+b_n \sqrt{2}) (1+\sqrt{2}) \\ a_{n+1}=a_n +2 b_n \\ b_{n+1} =a_n+b_n \end{eqnarray*} $(a_n,b_n)$ can be shown to satisfy Pell's equation inductively, as follows \begin{eqnarray*} a_{n+1}^2-2b_{n+1}^2 =a_n^2+4a_n b_n + 4b_n ^2 -2(a_n^2+2 a_n b_n +b_n^2) =-(a_n^2-2b_n^2) =(-1)^{n+1}. \end{eqnarray*}
Donald Splutterwit
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I actually did that but the problem is showing that the sequence gives all the solutions for pells equation . – Rz_CHIFFRE Dec 25 '17 at 15:16