$ℤ_n[i]$ is integral domain?

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对哪些$n\in\{2,3,\dots\}$, $x^2+1\pmod n$可约?
例如$n=2$有$(x-1)(x-1)\equiv x^2+1\pmod 2$

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$(x-a)(x+a)\equiv x^2+1\pmod n\iff n\mid a^2+1\iff n\in\tt\href{https://oeis.org/A008784}{A008784}$
其至多有一个因数 2, 奇素因数全$\equiv1\pmod4$, 即
$$n=2p_1^{\alpha_1}\cdots p_n^{\alpha_n},\quad p_i素数,\quad p_i\equiv1\pmod4$$

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For which $n$, $ℤ_n[x]/⟨x^2+1⟩$ is an integral domain?
Because $ℤ_n$ is a subring of $ℤ_n[x]$, we need $n$ to be prime number.
$\exists a:n\mid a^2+1\implies n\equiv1\pmod4\implies n\in\tt\href{https://oeis.org/A002144}{A002144}$
There is a Wikipedia entry Pythagorean prime

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Wikipedia Representation as a sum of two squares

Fermat's theorem on sums of two squares states that the prime numbers that can be represented as sums of two squares are exactly 2 and the odd primes congruent to 1 mod 4. The representation of each such number is unique, up to the ordering of the two squares.

PlanetMath sums of two squares

\begin{align*}(a^2+b^2)(c^2+d^2)&=(ac-bd)^2+(ad+bc)^2\\&=(ac+bd)^2+(ad-bc)^2\end{align*}
Thus in most cases, we can get two different nontrivial sum forms (i.e. without a zero addend) for a given product of two sums of squares.  For example, the product $65=5⋅13=(2^2+1^2)(3^2+2^2)$ attains the two forms $4^2+7^2$ and $8^2+1^2$.