I'm a high school student trying to understand something that may be ever so slightly beyond my reach. On the paper here: http://www.rose-hulman.edu/mathjournal/archives/2003/vol4-n1/paper3/v4n1-3pd.pdf, I've understood everything before Theorem 1.2, and I've understood the first half of Lemma 1.1 after a bit of research. (It doesn't seem like I need the second half, but I might be wrong.)
I thought I understood Theorem 1.2 at first, but it turns out I don't. The proof by contradiction seemed to make sense, but couldn't you arrive at the exact same contradiction by assuming that p is congruent to 1 mod 4? I don't see anything in the proof that utilizes the fact that p is a Gaussian prime.
Like I said earlier, I'm only a high school student, so feel free to dumb down your answer. Thanks for your help!
EDIT: Someone suggested I post a screenshot. Here it is: https://i.stack.imgur.com/JAP0A.jpg.
Let $R$ be an integral domain (e.g. $\mathbb{Z}$), and let $a, b \in R$. We say that $a$ divides $b$ in $R$ if there exists a $k \in R$ such that $ak = b$. If $p \in R$, we say that $p$ is prime in $R$ if whenever $p$ divides a product $ab$ of elements $a, b \in R$, then $p$ divides at least one of the factors $a$ or $b$. If $S$ is a larger integral domain containing $R$, and $p$ is a prime element of $R$, then $p$ may or may not be prime when considered as an element of $S$.
In Theorem 1.2, you have the equation $2e^2 = a^2 + s^2 = (a+si)(a-si)$.
If $p$ is a prime which is congruent to $3$ modulo $4$, then it is still prime as an element of $\mathbb{Z}[i]$. Since $p$ divides the product $(a+si)(a-si)$, it must divide one of the factors $a+si$ or $a-si$.
However, if $p$ is congruent to one, then $p$ is no longer prime as an element of $\mathbb{Z}[i]$. So the argument about dividing one of the factors no longer works.