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i have some confusion not getting in my head

this answer :Euclidean domain $\mathbb{Z}[\sqrt{d}]$

My confusion :enter image description here

My attempt : if i take $d= 10$ then $\mathbb{Z}[\sqrt 10]$ is not euclidean domain but it satisfied $10 \equiv 2 \mod 4$

similarly if i take $d= 13$ then $\mathbb{Z}[\sqrt 13]$ is euclidean domain but $13 \equiv 1 \mod 4$ which is contradicts to above answer

Im confused why he take $ d \equiv 2, 3 \mod 4 $ is euclidean domain ?

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    $\begingroup$ $R=\mathbb{Z}[\sqrt{13}]$ is not a Euclidean domain, because $\alpha=\frac{1+\sqrt{13}}{2} \notin R$, but $\alpha \in Frac(R)$, and $\alpha^2=3+\alpha$, so $R$ is not normal. On the other hand, the page gives you only a necessary condition, meaning that if $\mathbb{Z}[d^{1/2}]$ is an Euclidean domain, then $d-2$ or $d-3$ must be divisible by $4$, but the converse is not necessarily true. $\endgroup$ – Mindlack Aug 26 at 9:50
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A necessary condition for a subring of a number field to be an Euclidean domain is to be integrally closed. If $d\equiv 1 \pmod{4}$, then the ring $\mathbb{Z}[\sqrt{d}]$ is not integrally closed, since $\alpha=\frac{1+\sqrt{d}}{2}\in \mathbb{Q}(\sqrt{d})$ verifies that $$\alpha^2=\alpha+\frac{d-1}4,$$ so $\alpha$ is a root of $$X^2-X+\frac{1-d}4\in \mathbb{Z}[X],$$ but $\alpha$ is not in $\mathbb{Z}[\sqrt{d}]$.

This is the reason that you need $d\equiv 2,3\pmod{4}$ for the ring $\mathbb{Z}[X]$ to be an Euclidean domain. Note, however, that this condition is not sufficient.

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