<s>[INST] Which rational primes less than 50 are also Gaussian primes? Which rational primes less than 50 are also Gaussian primes?
My attempt: First we need to list all of the rational prime numbers that are less than $50$
$1,2,3,5,7,11,13,17,19,23,29,31,37,39,41,43,47$
A Gaussian prime is defined as a Gaussian integer that is not the product of Gaussian integers of similar norm.
The $Z[i]$ norm is $norm(a+bi)=\mid a+bi \mid^2 = a^2+b^2$ We have to look at the norms to determine which rational prime is a Gaussian prime. 
For example $4+i$ is a Gaussian prime because our norm is $norm(4+i)=4^2+(i)^2=4^2+(-1)^2=16+1=17$. Since $17$ is also a prime in a set of all integers, $4+i$ is not the product of Gaussian integers of smaller norm because no such norms divide 17.
Also we need to separate the prime factors into two groups: any prime factors congruent to 3 mod 4 and any prime factors congruent to 1 mod 4
So all of the numbers congruent to 3 mod 4 which is also in $4k+3$ format are $3,7,11,19,23,31,39,43$ and $47$. These numbers can't be expressed as the sum of two squares. 
All of the numbers congruent to 1 mod 4 which is also $4k+1$ are $1,5,13,17,29,37$, and $41,$ so these numbers can be expressed as the sum of squares and can be expressed as Gaussian Integers as well. 
$1=1^2+0^2 = (1+0i)(1-0i)$
$5=2^2+1^2=(2+i)(2-i)$
$13 =2^2+3^2=(2+3i)(2-3i)$
$17=4^2+1^2=(4+i)(4-i)$
$29=5^2+2^2=(5+2i)(5-2i)$
$37=6^2+1^2=(6+i)(6-i)$
$41=5^2+4^2=(5+4i)(5-4i)$
Now my problem is...how do I really know that these numbers which are represented as a sum of squares and Gaussian Integers is really a Gaussian Prime. I know 17 counts as one, and for some reason 2 is not a Gaussian Prime because $norm(1+i)=1+(i^2)=1+((-1)^2)=1+1=2$ is smaller than $norm(2^2)=4$. So the norm has to be the same value in order to be considered as a Gaussian Prime?
 [/INST] What might really help you understand here is the concept of conjugate pairs. The conjugate of $a - bi$ is $a + bi$. And $(a - bi)(a + bi) = a^2 + b^2$, which should look very familiar because it's the norm function.
Given $p$ a "rational prime" (for lack of a better term) with nonzero real part but no imaginary part, its norm is $p^2$. But if $p = a^2 + b^2$, then it is composite in $\mathbb{Z}[i]$ and its factorization is $(a - bi)(a + bi)$. It is a very well known result that the only odd rational primes that are the sum of two squares are primes of the form $4k + 1$. For example, $13 = 3^2 + 2^2 = (3 - 2i)(3 + 2i)$, so $3 + 2i$ is a Gaussian prime but $13$ is not.
Thus, minding the special case $(1 - i)(1 + i) = 2$, to find the rational primes between $1$ and $50$ that are also Gaussian primes, we just take the rational primes in that range, remove $2$ and remove the rational primes of the form $4k + 1$. So we take $$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$$ and whittle it down to $$3, 7, 11, 19, 23, 31, 43, 47.$$
The reason we can be absolutely sure that these really are Gaussian primes is by verifying their divisors. For example, what are the divisors of $3$ in $\mathbb{Z}[i]$? We only need to look at Gaussian integers with a norm of $9$ or less to find: $1, i, -1, -i, 3, 3i, -3, -3i$. These are just units and $3$ multiplied by units. Gaussian primes have exactly $8$ divisors.
As for $39$, it can't be expressed as a sum of two squares, but it is a composite number. If it's composite in $\mathbb{Z}$, it sure as heck is also composite in $\mathbb{Z}[i]$. Its factorization is $39 = 3(3 - 2i)(3 + 2i)$.
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