I'm surprised no one's been a jerk to you about listing 1 and 39 with the rational primes. There ought to be some civility in pointing out basic facts like that. The inclusion of 39 must have been a simple slip of the finger.
But the point still needs to be made that 1 is not a prime number, it is a unit. In an imaginary quadratic ring like $\mathbb{Z}[i]$, if $u$ is a unit, then $N(nu) = N(u)$. The units in $\mathbb{Z}[i]$ are $1, -1, i, -i$. These are not primes. So our list of potential Gaussian primes with real part but without imaginary part is therefore:
$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$
Your definition of "Gaussian prime" is at best confusing and at worst imprecise. Let me give you a couple of definitions that you can carry over to other quadratic integer rings:
- In $\mathbb{Z}[\sqrt{d}]$, the norm function is $N(a + b\sqrt{d}) = a^2 - db^2$. (If $d = -1$, this works out to $a^2 + b^2$). Also, the norm function is completely multiplicative, meaning that $N(mn) = N(m)N(n)$.
- A number $p = a + b\sqrt{d}$ is irreducible if for every possible way of expressing it as $p = \alpha \beta$ (with both $\alpha$ and $\beta$ being numbers in $\mathbb{Z}[\sqrt{d}]$ it turns out that either either $\alpha$ or $\beta$ is a unit (but not both).
- If $\mathbb{Z}[\sqrt{d}]$ is a unique factorization domain (as $\mathbb{Z}[i]$ is), then all irreducibles in that domain are prime numbers.
So we're looking to see which of the numbers from our list of potential primes are indeed irreducible in the domain of Gaussian integers and are therefore Gaussian primes. The norm function in $\mathbb{Z}[i]$ is $N = a^2 + b^2$. This means that we're essentially looking for prime numbers that can't be expressed as a sum of two squares.
I'm no historian, I don't know if Fermat proved this himself, but it's attributed to Fermat the theorem that if $p = a^2 + b^2$, then either $p = 2$ or $p \equiv 1 \pmod 4$. So then all we have to do is strike 2 and primes of the form $4k + 1$ off our list of Gaussian primes:
$3, 7, 11, 19, 23, 31, 43, 47$
If you still have doubts this is correct, you can put this sequence into the OEIS as a search. You should get Sloane's A002145 as a result, which includes the comment Natural primes which are also Gaussian primes.
I'm not going to get on your case about the supposed futility of asking for primes $p < 50$; instead I'm going to assume that you meant to ask for primes $p$ such that $N(p) < 50$. If you put your compass center at 0 on the complex plane and your compass pencil at 50, then draw the circle, you're essentially asking for all the primes within that circle. Our list of primes is quite incomplete at this point.
But we can easily quadruple that list with simple multiplication by units. That is,
$-3, -7, -11, -19, -23, -31, -43, -47$
$3i, 7i, 11i, 19i, 23i, 31i, 43i, 47i$
$-3i, -7i, -11i, -19i, -23i, -31i, -43i, -47i$
are all Gaussian primes! But we're still missing primes that have both a real and an imaginary part.
I hope you haven't forgotten about those natural primes that are Gaussian composites. If a natural prime $p = 2$ or $p \equiv 1 \pmod 4$, it can be expressed as $p = (a - bi)(a + bi)$, so $p$ is not a Gaussian prime but both $a - bi$ and $a + bi$ are Gaussian primes (this is probably a theorem, lemma or corollary in your book so I won't give a proof). Therefore we're now looking for Gaussian integers with nonzero real part $a$ and nonzero imaginary part $bi$ so that $a^2 + b^2 = p$, a natural prime. Since there is only one even natural prime (2), in almost all cases $a$ has to be odd and $b$ even or vice-versa, and in only four cases can both $a$ and $b$ be odd.
Limiting ourselves to $0 < a < 50$ and $0 < b < 50$, we obtain this list:
$1 + i$, $2 + i$, $3 + 2i$, $4 + i$, $5 + 2i$, $5 + 4i$, $6 + i$
Now simple multiplication by units will give you the other three lists necessary to complete your full list of Gaussian primes with norm less than 50.
To review: if $p$ is a Gaussian prime and $\Re(p) = 0$ or $\Im(p) = 0$, then $N(p)$ is a perfect square. But if $\Re(p) \neq 0$ and $\Im(p) \neq 0$, then $N(p)$ is a natural prime.
I think I know what you meant by "rational prime" and there are some authors who use it to mean the same thing; I find the term problematic for $\mathbb{Z}[i]$ but that's a whole other can of worms.