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I'm trying to show that $\mathbb{Z}[\sqrt{11}]$ is Euclidean with respect to the function $a+b\sqrt{11} \mapsto|N(a+b\sqrt{11})| = | a^2 -11b^2|$

By multiplicativity, it suffices to show that $\forall x \in \mathbb{Q}(\sqrt{11}) \exists n \in \mathbb{Z}(\sqrt{11}):|N(n-x)| < 1$

For the analogous statement for $\mathbb Z [\sqrt6]$, it worked by considering different cases, so I tried to do the same thing here. Here is what I did so far:

Let $x+y\sqrt{11} \in \mathbb Q (\sqrt{11})$

Case 1: Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \frac{1}{\sqrt{11}}$, then we can choose such a $b$ and a $a \in \mathbb Z$ s.t. $|x-a| \leq \frac{1}{2}$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$

From now on suppose $\forall b \in \mathbb Z: |y-b| > \frac{1}{\sqrt{11}}$

Case 2: Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \sqrt{\frac{5}{44}}$ Then we have $1 < 11 (y-b)^2 < \frac{5}{4}$, so we can choose $a \in \mathbb Z$ such that $\frac{1}{2} \leq |x-a| \leq 1$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$

From now on suppose $\forall b \in \mathbb Z: |y-b| > \sqrt{\frac{5}{44}}$

Case 3: Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \sqrt{\frac{2}{11}}$ Then we can choose $a \in \mathbb Z $ s.t. $1 \leq |x-a| \leq \frac{3}{2}$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$

From now on, we may suppose that $|y-b| > \sqrt{\frac{2}{11}}$.

This is where I'm stuck. I tried choosing $b \in \mathbb Z$ s.t. $\frac{1}{2} \geq |y-b| > \sqrt{\frac{2}{11}}$, but then I run into problems, whether I choose $a \in \mathbb Z$ s.t. $1 \leq |x-a| \leq \frac{3}{2}$ or s.t. $ \frac{3}{2} \leq |x-a| \leq 2$

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  • 1
    $\begingroup$ Look at math.stackexchange.com/questions/124484/… It appears that 11 will require at least 12, maybe 16 "starfish" $\endgroup$
    – Will Jagy
    Apr 3 '17 at 0:04
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    $\begingroup$ same history as 6, proved first by Perron in 1932, then Oppenheim 1934. $\endgroup$
    – Will Jagy
    Apr 3 '17 at 0:07
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I believe the result is proved in Oppenheim, Quadratic fields with and without Euclid's algorithm, Math Annalen 109 (1934) 349-352, and I think this paper is freely available here. The proof is essentially the first half of page 350, together with preliminary observations on page 349.

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Division algorithm for Euclidean domain $\mathbb{Z}[\sqrt{11}]$

While Oppenheim (1934) proved that there is a division algorithm for $\mathbb{Z}[\sqrt{11}]$. No examples was given explicitly. Inspired by his proof, one suitable division algorithm was identified. Hence, one could compute in $\mathbb{Z}[\sqrt{11}]$.

Let $R = \mathbb{Z}[\sqrt{11}] = \{ s+t\sqrt{11} : s,t \in \mathbb{Z} \} $ be a quadratic integer ring. Let $\alpha = s+t\sqrt{11} \in R$. Define norm $N: R \setminus\{0\} \rightarrow \mathbb{Z}_{\geq 0}$ by setting $\alpha \mapsto |s^2-11t^2|$, which is well defined in $R$. Proof omitted. Norm $N$ is also well defined in $\mathbb{Q}[\sqrt{11}]$. Proof omitted. There exists $\alpha/\beta \in \mathbb{Q}[\sqrt{11}]$ $=$ $\{ s+t\sqrt{11} : s,t \in \mathbb{Q} \} \supset R $ with a numerator and a denominator $\alpha, \beta \in R$, $\beta \ne 0$. Proof omitted.

Suppose the nearest lattice point to point $\alpha / \beta$ is $m+n\sqrt{11} \in R$ (i.e. $m,n \in \mathbb{Z}$) with $\alpha / \beta = (m+n\sqrt{11}) + (a+b\sqrt{11})$, $a,b \in \mathbb{Q}$, $-\frac{1}{2} \leq a \leq \frac{1}{2}$ and $-\frac{1}{2} \leq b \leq \frac{1}{2}$. We will consider lattice point $(m,n) = m+n\sqrt{11}$ and five other nearby lattice points. They, $(m+x,n+y)$ with $(x,y) \in \{ (0,0), (1,0), (-1,0), (2,0), (2,1), (-5,2) \} $, are candidate quotients $q$ to a division algorithm. By considering only positive values of $a,b$ such that $0 \leq a \leq \frac{1}{2}$ and $0 \leq b \leq \frac{1}{2}$, similar arguments could be made for the three other quadrants by symmetry. With $q=(m+x,n+y)$, rewrite $\alpha = q \beta +r$ as:

\begin{align} \alpha / \beta &= q + r/\beta \\ &= \big( (m+n\sqrt{11}) + (x+y\sqrt{11}) \big) + \big( (a+b\sqrt{11}) - (x+y\sqrt{11}) \big) \end{align}

Inequality $N(r) < N(\beta)$ is true $\iff$ $N(r)/N(\beta) < 1$ $\iff$ $N(r/\beta) < 1$ $\iff$ $N \big( (a+b\sqrt{11}) - (x+y\sqrt{11}) \big)$ $= |(x-a)^2 - 11(y-b)^2| < 1$ . Proof omitted.

Case 1: Consider lattice point $(m,n)$, $0 \leq a^2 \leq \frac{1}{4}$ and $0 \leq 11b^2 < 1$. The maximum possible difference occurs when $a^2 = 0$ and $11b^2 = 1^-$. Therefore $|a^2 - 11b^2| < 1$ and we have:

\begin{align} 0 &\leq a \leq \frac{1}{2} \\ 0 &\leq b < \sqrt{ \frac{1}{11} } \approx 0.302 \end{align}

seven cases bar chart

Case 2: Consider lattice point $(m+1,\ n)$, $\frac{1}{4} \leq (1-a)^2 \leq 1$ and $1 < 11b^2 < \frac{5}{4}$. The maximum possible difference occurs when $(1-a)^2 = \frac{1}{4}$ and $11b^2 = \frac{5}{4}^-$. Therefore $|(1-a)^2 - 11b^2| < 1$ and we have:

\begin{align} 0 &\leq a \leq \frac{1}{2} \\ \sqrt{ \frac{1}{11} } &< b < \sqrt{ \frac{5}{44} } \approx 0.337 \end{align}

Case 3: Consider lattice point $(m-1,\ n)$, $1 \leq (1+a)^2 \leq \frac{9}{4}$ and $\frac{5}{4} < 11b^2 < 2$. The maximum possible differences are $2^- - 1 < 1$ and $\frac{9}{4} - \frac{5}{4}^+ < 1$. Therefore $|(1+a)^2 - 11b^2| < 1$ and we have:

\begin{align} 0 &\leq a \leq \frac{1}{2} \\ \sqrt{ \frac{5}{44} } &< b < \sqrt{ \frac{2}{11} } \approx 0.426 \end{align}

Case 4: Consider lattice point $(m+2,\ n)$, $\frac{9}{4} \leq (2-a)^2 < 3$ and $2 < 11b^2 \leq \frac{11}{4}$. The maximum possible difference occurs when $(2-a)^2 = 3^-$ and $11b^2 = 2^+$. Therefore $|(2-a)^2 - 11b^2| < 1$ and we have:

\begin{align} 0.270 \approx 2-\sqrt{3} &< a \leq \frac{1}{2} \\ \sqrt{ \frac{2}{11} } &< b \leq \frac{1}{2} \end{align}

Case 5: Consider lattice point $(m+2,\ n+1)$, $3 < (2-a)^2 \leq 4$ and $3 < 11(1-b)^2 < 4$. The maximum possible difference are $4^- - 3^+ < 1$ and $4 - 3^+ < 1$. Therefore $|(2-a)^2 - 11(1-b)^2| < 1$ and we have:

\begin{align} 0 &\leq a < 2-\sqrt{3} \\ 0.397 \approx 1-\sqrt{ \frac{4}{11} } &< b < 1-\sqrt{ \frac{3}{11} } \approx 0.478 \end{align}

Case 6: Consider lattice point $(m+2,\ n+1)$ again with different ranges of $a,b$, $\frac{9}{4} \leq (2-a)^2 < \frac{15}{4}$ and $\frac{11}{4} \leq 11(1-b)^2 < \frac{13}{4}$. The maximum possible differences are $\frac{13}{4}^- - \frac{9}{4} < 1$ and $\frac{15}{4}^- - \frac{11}{4} < 1$. Therefore $|(2-a)^2 - 11(1-b)^2| < 1$ and we have:

\begin{align} 0.064 \approx 2 - \sqrt{ \frac{15}{4} } &< a \leq \frac{1}{2} \\ 0.456 \approx 1-\sqrt{\frac{13}{44}} &< b \leq \frac{1}{2} \end{align}

Case 7: Consider lattice point $(m-5,\ n+2)$, $25 \leq (5+a)^2 < \frac{103}{4}$ and $\frac{99}{4} \leq 11(2-b)^2 < \frac{103}{4}$. The maximum possible difference occurs when $(5+a)^2 = \frac{103}{4}^-$ and $11(2-b)^2 = \frac{99}{4}$. Therefore $|(5+a)^2 - 11(2-b)^2| < 1$ and we have:

\begin{align} 0 &\leq a < \sqrt{ \frac{103}{4} } - 5 \approx 0.074 \\ 0.470 \approx 2 - \sqrt{\frac{103}{44}} &< b \leq \frac{1}{2} \end{align}

All boundary cases are ignored, since $\sqrt{ \frac{1}{11} }$, $\sqrt{ \frac{5}{44} }$, $1-\sqrt{ \frac{4}{11} }$, $\sqrt{ \frac{2}{11} }$, $1-\sqrt{\frac{13}{44}}$, $2 - \sqrt{\frac{103}{44}}$, $1-\sqrt{ \frac{3}{11} }$, $2 - \sqrt{ \frac{15}{4} }$, $\sqrt{ \frac{103}{4} } - 5$, $2-\sqrt{3}$ $\notin \mathbb{Q}$.

Since the quadratic integer ring $\mathbb{Z}[\sqrt{11}]$ has (a) a well defined norm and (b) a division algorithm $\alpha = q \beta + r$ such that either $r=0$ or $N(r)<(b)$, for all numerators and denominators $\alpha, \beta \in \mathbb{Z}[\sqrt{11}]$, $\beta \ne 0$ with some quotient and remainder $q, r \in \mathbb{Z}[\sqrt{11}]$, it is a Euclidean domain. $\blacksquare$

different ranges of a,b

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  • $\begingroup$ Example, with $a=0.070$ and $b=0.475$, let $\alpha = -4992+1637\sqrt{11}$ and $\beta = 1400-400\sqrt{11}$. A division algorithm $-4992+1637\sqrt{11} = \left(1400-400\sqrt{11}\right)\left(3+2\sqrt{11}\right)-392+37\sqrt{11}$ with $138605 = N \left( -392+37\sqrt{11} \right) < N \left( \beta \right) = 200000$. Another division algorithm $-4992+1637\sqrt{11}= \left(1400-400\sqrt{11}\right)\left(-4+3\sqrt{11}\right)+13808-4163\sqrt{11}$ with $24605 = N \left( 13808-4163\sqrt{11} \right) < N \left( \beta \right) = 200000$. $\endgroup$ Feb 3 at 23:28
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I am a novice and learning to write simple proofs. I welcome corrections and suggestions. Below is a detailed workout. Most maths students will find it unnecessarily verbose. Oppenheim (1934) proved, among other proofs, that $\mathbb{Z}[\sqrt{11}]$ has a division algorithm and hence a Euclidean domain. We unpack his proof here.

Let $R = \mathbb{Z}[\sqrt{11}] = \{ s+t\sqrt{11} : s,t \in \mathbb{Z} \} $ be a quadratic integer ring. It is said to be a Euclidean domain if there are:

(a) a well defined norm $N(\alpha) \leq N(\alpha\beta)$, and

(b) a division algorithm $\alpha = q \beta +r$ such that either (i) $r=0$, or (ii) $r \ne 0$ and $N(r) < N(\beta)$

with $\alpha,\beta \in R$, $\beta \ne 0$ and some quotient and remainder $q,r \in R$.

Let $\alpha = s+t\sqrt{11} \in R$. Define norm $N: R \setminus\{0\} \rightarrow \mathbb{Z}_{\geq 0}$ by setting $\alpha \mapsto |s^2-11t^2|$, which is well defined in $R$. Proof omitted. Norm $N$ is also well defined in $\mathbb{Q}[\sqrt{11}]$. Proof omitted. There exists $\alpha/\beta \in \mathbb{Q}[\sqrt{11}] = \{ s+t\sqrt{11} : s,t \in \mathbb{Q} \} \supset R $ with a numerator and a denominator $\alpha, \beta \in R$. Proof omitted.

Suppose the nearest lattice point to point $\alpha / \beta$ is $m+n\sqrt{11} \in R$ (i.e. $m,n \in \mathbb{Z}$) with $\alpha / \beta = (m+n\sqrt{11}) + (a+b\sqrt{11})$, $a,b \in \mathbb{Q}$, $-\frac{1}{2} \leq a \leq \frac{1}{2}$ and $-\frac{1}{2} \leq b \leq \frac{1}{2}$. We will consider lattice point $(m,n) = m+n\sqrt{11}$ and five others nearby. They, $(m+x,n+y)$ with $(x,y) \in \{ (0,0), (1,0), (-1,0), (2,0), (2,1), (5,2) \} $, are candidate quotients to a division algorithm. By considering only positive values of $a,b$ such that $0 \leq a \leq \frac{1}{2}$ and $0 \leq b \leq \frac{1}{2}$, similar arguments could be made for the three other quadrants by symmetry. With $q=(m+x,n+y)$, rewrite $\alpha = q \beta +r$ as:

\begin{align} \alpha / \beta &= q + r/\beta \\ &= q + \big( (a+b\sqrt{11}) - (x+y\sqrt{11}) \big) \end{align}

Condition (b)(ii) if $r \ne 0$, $N(r) < N(\beta)$ is true $\iff$ $N(r)/N(\beta) < 1$ $\iff$ $N(r/\beta) < 1$ $\iff$ $N \big( (a+b\sqrt{11}) - (x+y\sqrt{11}) \big)$ $= |(x-a)^2 - 11(y-b)^2| < 1$. Proof omitted.

Proposition: A division algorithm holds for $\mathbb{Z}[\sqrt{11}]$ if both inequalities:

\begin{align} U(x,y): &\ (x-a)^2 - 11(y-b)^2 < 1 \ \text{and} \\ V(x,y): &\ 11(y-b)^2 - (x-a)^2 < 1 \ \text{are true,} \end{align}

for all $a,b \in \mathbb{Q}$, $0 \leq a \leq \frac{1}{2}$ and $0 \leq b \leq \frac{1}{2}$, and some $x,y \in \mathbb{Z}$.

Rewrite not $U(x,y)$ and not $V(x,y)$ as $P(x,y)$ and $N(x,y)$ respectively.

\begin{align} P(x,y): &\ (x-a)^2 \geq 1 + 11(y-b)^2 \\ N(x,y): &\ 11(y-b)^2 \geq 1 + (x-a)^2 \end{align}

Suppose to the contrary that the contrapositive of the above proposition is true. It asserts that a division algorithm does not hold for $\mathbb{Z}[\sqrt{11}]$. Then either $P(x,y)$ or $N(x,y)$ is true for some $a,b$ and all $x,y$.

Consider $\mathbf{P(0,0)} : (-a)^2 \geq 1 + 11(-b)^2$. It implies $a^2 \geq 1$. Take the positive root and we have $a \geq 1$ which contradicts $0 \leq a \leq \frac{1}{2}$. Since $P(0,0)$ is false, $N(0,0)$ must hold.

$N(0,0): 11b^2 \geq 1+a^2$

Consider $\mathbf{P(1,0)} : (1-a)^2 \geq 1 + 11b^2$. By $N(0,0)$,

\begin{align} (1-a)^2 &\geq 1 + 11b^2 \geq 2+a^2 \\ (1-a)^2 &\geq 2+a^2 \\ 1-2a+a^2 &\geq 2+a^2 \\ -2a &\geq 1 \\ a &\leq -\frac{1}{2} \end{align}

which contradicts $0 \leq a \leq \frac{1}{2}$. Since $P(1,0)$ is false, $N(1,0)$ must hold.

$N(1,0): 11b^2 \geq 1 + (1-a)^2$

a starfish and six candidate quotients q https://www.desmos.com/calculator/rhrza9vqpx https://www.desmos.com/calculator/pccntyashy

Consider $\mathbf{P(-1,0)} : (-1-a)^2 \geq 1 + 11b^2$. By $N(1,0)$,

\begin{align} (1+a)^2 &\geq 1 + 11b^2 \\ (1+a)^2 - 1 &\geq 11b^2 \geq 1 + (1-a)^2 \\ 1 + 2a + a^2 - 1 &\geq 1 + 1 - 2a + a^2 \\ 4a &\geq 2 \\ a &\geq \frac{1}{2} \\ \end{align}

Then $a = \frac{1}{2}$ since $0 \leq a \leq \frac{1}{2}$. It follows that:

\begin{align} (1+\frac{1}{2})^2 - 1 &\geq 11b^2 \geq 1 + (1-\frac{1}{2})^2 \\ \frac{5}{4} &\geq 11b^2 \geq \frac{5}{4} \end{align}

But $b = \sqrt{ \frac{5}{44} } \notin \mathbb{Q}$ contradicts $b \in \mathbb{Q}$. Since $P(-1,0)$ is false, $N(-1,0)$ must hold.

$N(-1,0): 11b^2 \geq 1 + (1+a)^2$

Consider $\mathbf{N(2,0)} : 11b^2 \geq 1 + (2-a)^2$.

\begin{align} 11b^2 &\geq 1 + (2-a)^2 \geq \frac{13}{4} > \frac{11}{4} \\ b^2 &> \frac{1}{4} \end{align}

Take the positive root and we have $b > \frac{1}{2}$. But it contradicts $0 \leq b \leq \frac{1}{2}$. Since $N(2,0)$ is false, $P(2,0)$ must hold.

$P(2,0): (2-a)^2 \geq 1 + 11b^2$

Consider $\mathbf{N(2,1)} : 11(1-b)^2 \geq (2-a)^2$. By $N(2,0)$,

\begin{align} 11(1-b)^2 &\geq 1 + (2-a)^2 \geq 1 + 11b^2 \\ 11 - 22b + 11b^2 &\geq 1 + 11b^2 \\ 9 &\geq 22b \\ b &\leq \frac{9}{22} \end{align}

By $N(-1,0)$,

\begin{align} 11b^2 &\geq 1 + (1+a)^2 \geq 2 > \frac{81}{44} \\ $b^2 &> (\frac{9}{22})^2 \end{align}

Take the positive root and we have $b > \frac{9}{22}$. But it contradicts $b \leq \frac{9}{22}$. Since $N(2,1)$ is false, $P(2,1)$ must hold.

$P(2,1): (2-a)^2 \geq 1 + 11(1-b)^2$

By $P(2,1)$ and $N(-1,0)$, they imply:

\begin{align} (2-a)^2 &\geq 1 + 11(1-b)^2 \\ (2-a)^2 &\geq 1 + 11 -22b + 11b^2 \\ (2-a)^2 -12 + 22b &\geq 11b^2 \geq 1 + (1+a)^2 \\ 4 -4a + a^2 -12 + 22b &\geq 2 + 2a + a^2 \\ 22b &\geq 10 + 6a \\ 11b &\geq 5 + 3a \end{align}

Out of interest. Take the least possible value to the RHS; it implies $11b \geq 5$. Likewise, take the greatest possible value to the LHS; it implies $11 \geq 10 + 6a$ $\implies$ $1 \geq 6a$.

Consider $\mathbf{N(5,2)}: 11(2-b)^2 \geq 1 + (5-a)^2$. By $P(2,1)$, it implies:

\begin{align} 11(2-b)^2 &\geq 1 + (5-a)^2 \\ 11(2-b)^2 &\geq 26 -10a +a^2 \\ 11(2-b)^2 &\geq (4 - 4a + a^2) + 22 -6a \\ 11(2-b)^2 - 22 + 6a &\geq (2-a)^2 \geq 1 + 11(1-b)^2 \\ 22 - 44b + 11b^2 + 6a &\geq (2-a)^2 \geq 12 -22b + 11b^2 \\ 10 + 6a &\geq 22b \\ 5 + 3a &\geq 11b \end{align}

By $11b \geq 5 + 3a$ and $5 + 3a \geq 11b$, we have $5 + 3a = 11b$. Retrace backwards the argument leading to $5 + 3a \geq 11b$. We have $11(2-b)^2 - 22 + 6a = (2-a)^2 = 1 + 11(1-b)^2$. Take the middle and the RHS of this chain of equality and name it $P'(2,1)$. Establishing the following two equalities are not strictly necessary for this proof. Since Oppenheim (1934) mentioned them, we will include them here. Take the LHS and the middle of the above chain of equality, we have $11(2-b)^2 - 22 + 6a = (2-a)^2$ $\implies$ $11(2-b)^2 = (4 - 4a + a^2) + 22 -6a$ $\implies$ $11(2-b)^2 = 26 -10a +a^2$ $\implies$ $11(2-b)^2 = 1 + (5-a)^2$. Name it $N'(5,2)$. Retrace backwards the argument leading to $11b \geq 5 + 3a$. We have $(2-a)^2 -12 + 22b = 11b^2 = 1 + (1+a)^2$. Take the middle and the RHS of this chain of equality and name it $N'(-1,0)$. And we have:

\begin{align} P'(2,1)&: (2-a)^2 = 1 + 11(1-b)^2 \\ N'(5,2)&: 11(2-b)^2 = 1 + (5-a)^2 \\ N'(-1,0)&: 11b^2 = 1 + (1+a)^2 \end{align}

By $N'(-1,0)$ and $5 + 3a = 11b$,

\begin{align} 11b^2 &= 1 + (1+a)^2 \\ (11b)^2 &= 11 + 11(1+a)^2 \\ (5+3a)^2 &= 11 + 11(1+a)^2 \\ 25 + 30a + 9a^2 &= 22 + 22a + 11a^2 \\ 2a^2 - 8a &= 3 > 0 \end{align}

It implies $a<0$ or $a > 4 > \frac{1}{2}$ but they contradict $0 \leq a \leq \frac{1}{2}$. And hence $N(5,2)$ is false, $P(5,2)$ must be true.

$P(5,2): (5-a)^2 \geq 1 + 11(2-b)^2$

Consider $\mathbf{P(5,2)}$, by $N'(-1,0)$, it implies:

\begin{align} (5-a)^2 &\geq 1 + 11(2-b)^2 \\ 25 - 10a + a^2 &\geq 1 + 11(2-b)^2 \\ 23 - 12a + 1 + (1 + 2a + a^2) &\geq 1 + 11(2-b)^2 \\ 23 - 12a + \big( 1 + (1 + a)^2 \big) &\geq 1 + 11(2-b)^2 \\ 23 - 12a + 11b^2 &\geq 45 - 44b + 11b^2 \\ 44b &\geq 22 +12a \\ 22b &\geq 11 + 6a \end{align}

Possible range of the LHS is $[0,11]$ and that of RHS is $[11,14]$. Therefore, they only possible values are $b = \frac{1}{2}$ and $a=0$. Substitute $b = \frac{1}{2}$ back into $P(5,2)$, and it asserts,

$(5-a)^2 \geq 1 + 11(2- \frac{1}{2} )^2 = \frac{103}{4} > (\frac{10}{2})^2 = 5^2$

Take the positive root (since negative root $a-5 \geq 0$ is false) and we have $5-a > 5$ $\implies$ $a < 0$. But it contradicts $a = 0$. And hence $P(5,2)$ is false.

Since both $N(5,2)$ and $P(5,2)$ are false, they contradicts the proposition that either $P(x,y)$ or $N(x,y)$ is true for some $a,b$ and all $x,y$. And hence, the proposition that a division algorithm does not hold is false. Therefore, $\mathbb{Z}[\sqrt{11}]$ has a division algorithm and hence a Euclidean domain. $\ \ \ \ \blacksquare$

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Another way to graph it for showing that $\mathbb{Z}[{\sqrt{11}}]$ is norm-euclidean. However, this method fails for $\mathbb{Z}[{\sqrt{19}}]$. There are a few small gaps that this elementary method fails to cover.

Z[sqrt(11)] with max a,b ranges
Z[sqrt(11)] with max a,b ranges

(cases 1,5,6)
Z[sqrt(11)] with max a,b ranges (cases 1,5,6)

(cases 2,4)
Z[sqrt(11)] with max a,b ranges (cases 2,4)

(case 3)
Z[sqrt(11)] with max a,b ranges (case 3)

(case 7)
Z[sqrt(11)] with max a,b ranges (case 7)

(areas covered by only one algorithm)
Z[sqrt(11)] with max a,b ranges (only alg. only)

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  • $\begingroup$ Can you relate this to the other two answers you have posted? $\endgroup$ Jun 5 at 3:58
  • $\begingroup$ It is related to the former dated Feb 2 at 20:20 which identified one division algorithm for $\mathbb{Z}[\sqrt{11}]$. Among the new sketches, in each case, it shades off a maximum possible area instead of a rectangle patch. The motivation was an [failed] attempt to extend this method to $\mathbb{Z}[\sqrt{19}]$. There are a few small patches I couldn't covered with combinations of $x,y$ up to $\pm 50$. Apparently, with this elementary method, I can't (at least not easily) find a complete division algorithm for $\mathbb{Z}[\sqrt{19}]$ for all possible values of $a,b$. $\endgroup$ Jun 5 at 8:38

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