Proving [imath]10+3\frac{1+\sqrt{69}}{2}[/imath] is prime in [imath]\mathbb{Z}\frac{1+\sqrt{69}}{2}[/imath]

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Let's note [imath]\theta=10+3\frac{1+\sqrt{69}}{2}[/imath]. Suppose [imath]\theta \mid xy[/imath], with [imath]x, y \in \mathbb{Z}[\frac{1+\sqrt{69}}{2}][/imath]. It's pretty easy to see [imath]N(\theta)=23[/imath], so [imath]23\mid N(x)[/imath] or [imath]23\mid N(y)[/imath] (since 23 is prime in [imath]\mathbb{Z}[/imath]).

I have trouble now proving [imath]\theta \mid x[/imath]. What I do have right now is that [imath]\pm 23=\theta \bar{\theta} \mid x \bar{x}[/imath]. Think it's possible to prove [imath]\theta[/imath] is prime, without using too many results from number theory?

 

[MathNugget]

Oops, seems I found the answer. For anyone interested: https://www.researchgate.net/public...ization_of_Z_1_-d_2_for_d_3_7_11_19_43_67_163

Apparently, it's fairly easy to use ideals and factorisations to prove any element x of [imath]\mathbb{Z}[\alpha][/imath], with [imath]\alpha[/imath] a root of an irreducible 2nd degree polynomial in [imath]\mathbb{Z}[X][/imath], that has the [imath]x\bar{x}=[/imath] a prime number, is prime in the ring.