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Let $R$ be the ring ${\mathbb Z}[\sqrt{2}]$. For $z\in R$, $z=x+y\sqrt{2}\in R$ with $x$ and $y$ in $\mathbb Z$, put $\|z\|={\sf max}(|x|,|y|)$ and

$$D_z=\left\{ (a,b) \in R^2 : ab=z,\ a \text{ and } b \text{ are not units in } R\right\}$$

and

$$ \rho(z)= \begin{cases} 0,& \text{ if} \ D_z=\emptyset; \\ {\sf min}\left(\|a\|,(a,b) \in D_z^2\right), & \text{ if} \ D_z\neq\emptyset. \end{cases} $$

Finally, set $\mu(m)={\sf max}(\rho(z) : \|z\| \leq m)$ for an integer $m$. Is anything at all known about the growth and the asymptotics of $\mu$ ? For example, are there any polynomial or exponential, upper or lower bounds ?

Update 09/10/2012 : It seems that there is a linear bound, namely $\rho(z) \leq ||z||$. To show this, it would suffice to show that when ${\sf gcd}(x,y)=1$ and $p$ is a prime divisor of $x^2-2y^2$, then we have a representation $p=a^2-2b^2$ with ${\sf max}(|a|,|b|) \leq {\sf max}(|x|,|y|)$.

The existence of $(a,b)$ (without the last inequality) is well-known, but the bound on $|a|$ and $|b|$ does not seem easy to establish.

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1 Answer 1

If $x$ is even in $z=x+y\sqrt 2$ then $z$ is a multiple of $\sqrt 2$, hence $\rho(z)\le 1$.

On the other hand, if $|N(z)|=|x^2-2y^2|$ is the square of a prime $p$ (and $x+y\sqrt 2$ is not a prime in $R$), then we must have $N(a+b\sqrt2)=a^2-2b^2=\pm p$, which implies $a^2\ge p$ or $b^2\ge \frac2p$, hence $||a+b\sqrt 2||\ge \sqrt{\frac p2}$ and fnally $\rho(z)\ge \sqrt{\frac p2}$.

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