Prime in $\mathbb Z [\sqrt{-5}]$ but not in $\mathbb Z [\sqrt{5}]$ 
Determine whether $\sqrt{-5}$ is irreducible and/or prime in $\mathbb Z[\sqrt{-5}]$.
What is a prime $p>5$ which is prime in $\mathbb Z [\sqrt{-5}]$ not prime in $\mathbb Z [\sqrt{5}]$?

For the first one, I believe $\sqrt{−5}$ is irreducible as $N(\sqrt{−5}) = 5$ and the only integers that divide $5$ are $1$ (where all elements with norm $1$ are units) and $5$ (where only $±\sqrt{−5}$ have norm $5$). Is this a good explanation? I'm guessing that $\sqrt{-5}$ is prime but I'm not sure how to justify why.
For the second one, I can't think of any primes over 5.
Help would be great!
 A: Yes, $\sqrt{-5}$ is irreducible and prime in $\mathbb{Z}[\sqrt{-5}]$. If it wasn't, it would be possible to find numbers in the domain such that $ab = c \sqrt{-5}$ yet $\sqrt{-5}$ divides neither $a$ nor $b$. Since the norm is multiplicative, that would require $N(ab) = 5$ and so the only possibilities are $N(a) = 1$, $N(b) = 5$ or vice-versa.
For the second part of your question, by $\mathbb{Z}[\sqrt{5}]$ do you mean $\mathbb{Z}[\phi]$ where $$\phi = \frac{1 + \sqrt{5}}{2}?$$ I'll assume that you do. Since $x^2 \equiv \pm 2 \pmod 5$ is insoluble in integers, we can be assured that primes ending in $3$ or $7$ are at least irreducible in both domains (and of course prime in $\mathbb{Z}[\phi]$).
So what we're looking for is that $x^2 \equiv -5 \pmod p$ has no solutions but $x^2 \equiv 5 \pmod p$ does. I'd work out the Legendre symbol for you but I'm running late to dinner.
A: The condition for $p$ to split in a quadratic field of discriminant $D$ is that
$$\left(\frac{D}{p}\right)=+1$$
The discriminant of $\mathbb{Z}(\sqrt{-5})$ is $D=-20$, and of $\mathbb{Z}(\sqrt{5})$ is $D=5$, thus you seek a $p$ such that 
$$\left(\frac{-20}{p}\right)=-1, \left(\frac{5}{p}\right)=+1$$
Now $$\left(\frac{5}{p}\right)=\left(\frac{p}{5}\right)$$ and this is $+1$ if and only $p\equiv \pm 1 (\mod 5)$.
In this case 
$$\left(\frac{-20}{p}\right)=\left(\frac{-1}{p}\right)$$
so $p$ must also satisfy 
$$\left(\frac{-1}{p}\right)=-1$$ or that 
$p\equiv 3 (\mod 4)$.
The smallest such prime is $p=11$.
And indeed, 
$$11=(4+\sqrt{5})(4-\sqrt{5})$$
and on the other had it is easy to see that 
$$a^2+5b^2=11$$ is impossible.
One can also see that $19$ is another such prime. Thus in fact the set of such primes are those of the forms $20n+11$ and $20n-1$.
