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Suppose $d$ be a square-free integer and $\mathbb{Z}[\sqrt{d}]$ is a unique factorization domain.
If $p$ is prime and $d$ is a square mod $p$, then prove that $p$ is reducible in $\mathbb{Z}[\sqrt{d}]$.

Assume that $p$ is prime and $d$ is a square mod $p$.
By this assumption I can only get that $d$ is quadratic residue of $p$.
But, how to show that $p = (u_1 + v_1\sqrt{d})(u_2 + v_2\sqrt{d})$ in somehow ?

xxxxxx
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  • $p$ divides $a^2-d=(a+\sqrt{d})(a-\sqrt{d})$ (for a suitable $a$), but doesn't divide any of the factors – user8268 Nov 04 '21 at 07:39
  • @user8286 Do you mean that since $d$ is a square mod $p$, then $d \equiv a^2 ~(\text{mod}~p)$, which implies that $a^2 - d \equiv 0 ~(\text{mod}~p)$, which implies that $p$ divides $a^2-d$ ? – xxxxxx Nov 04 '21 at 07:47
  • @user8286 Also, I don't understand why can't $p$ divides the factors of $a^2-d$ ? Is that becasue the factors of $a^2-d$ are not integers ? – xxxxxx Nov 04 '21 at 07:49
  • yes for the first question, and for the second just notice that $(a\pm\sqrt d)/p$ is not in your ring – user8268 Nov 04 '21 at 11:25

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