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Chapter 38 of Mazur and Stein's "Prime Numbers and the Riemann Hypothesis" asks us to prove that if a Gaussian integer is a prime Gaussian integer, then its norm can only be an "ordinary prime number, or the square of an ordinary prime number".

$$|G_{\text{prime}}|=p \text{ or } p^2 $$

I have researched other questions and answers on this site and via google but cannot find a simple explanation that I can understand.

I'd appreciate an answer that avoids too much terminology. For example, this answer is not clear enough for me as someone not formally trained in mathematics: Problem about Gaussian prime

Note: the system suggests this is answered here, but the answer doesn't meet the request here for a accessibility to those without formal mathematics training: The norm of a Gaussian prime is a prime or a square of a prime in $\mathbb{Z}$

Penelope
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    Let $\pi$ be a Gaussian prime. Its norm ${\rm N}(\pi) = \pi\overline{\pi}$ is a positive integer greater than $1$. Factor this into a product of ordinary prime numbers, say $p_1p_2\cdots p_r$. Since $\pi$ is a factor of ${\rm N}(\pi)$, $\pi$ is a factor of $p_1p_2\cdots p_r$. Thus, since $\pi$ is prime, $\pi$ is a factor of some $p_i$. Therefore the positive integer ${\rm N}(\pi)$ is a factor of ${\rm N}(p_i) = p_i^2$, so ${\rm N}(\pi)$ is $p_i$ or $p_i^2$ (the norm is bigger than $1$). – KCd Jul 30 '20 at 02:30
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    @KCd your comment should be an answer. It's complete and directly address the question. – CyclotomicField Jul 30 '20 at 02:35
  • @JCAA thanks but sadly I am unable to follow that suggested answer. – Penelope Jul 30 '20 at 03:15
  • Then you should ask questions to the person who answered that question. – markvs Jul 30 '20 at 03:16
  • @KCd - I am almost there with your explanation, but not quite. You say that since $\pi$ is prime, it is also a factor of some $p_i$. Why is this? Why can't $\pi$ be a factor of several of the primes in the factorisation of $N(\pi)$? If I assume you are correct, and $\pi$ is a factor of only one of those primes $p_i$, then why is $N(\pi)$ a factor of $N({p_i})$? – Penelope Jul 30 '20 at 03:22
  • It is irrelevant if $\pi$ is a factor of several $p_i$: where is it ever mentioned that it couldn't divide more than one? I don't know why you're focusing on "only one" when that has no role anywhere in what I wrote. In $\mathbf Z$, if $2 \mid a_1a_2\cdots a_r$ then $2 \mid a_i$ for some $i$. Can it divide more than one $a_i$? Of course, but who cares? It divides some $a_i$ (maybe more than one, whatever) and that's good enough. – KCd Jul 30 '20 at 04:27

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