Are there prime numbers not irreducible in $\mathcal{O}_{\mathbb{Q}(\sqrt{577})}$? All small (rational) primes I've looked at seem to be irreducible elements in the ring $\mathcal{O}_{\mathbb{Q}(\sqrt{577})}$. Often they're irreducible but not prime. And of course $577 = (\sqrt{577})^2$.
Looking at ideals, it's clear that $$\langle 2 \rangle = \left\langle 2, \frac{1}{2} + \frac{\sqrt{577}}{2} \right\rangle^2,$$ $$\langle 3 \rangle = \left\langle 3, \frac{1}{2} - \frac{\sqrt{577}}{2} \right\rangle \left\langle 3, \frac{1}{2} + \frac{\sqrt{577}}{2} \right\rangle,$$ $\langle 5 \rangle$ and $\langle 7 \rangle$ are prime, etc.
It's then not difficult to find failures of unique factorization such as $$12 = 2^2 \times 3 = (-1) \left(\frac{23}{2} - \frac{\sqrt{577}}{2} \right) \left(\frac{23}{2} + \frac{\sqrt{577}}{2} \right).$$ 
But I can't seem to find a case of principal ideals $\langle a \pm b \sqrt{577} \rangle$, with nonzero $a, b \in \mathbb{Q}$ (either integers or halves of integers, to be precise) having prime norm.
For contrast, observe that in $\mathbb{Z}[\sqrt{15}]$ (not UFD either), we have $\langle 11 \rangle = \langle 2 - \sqrt{15} \rangle \langle 2 + \sqrt{15} \rangle$.
Is this theoretically impossible in $\mathcal{O}_{\mathbb{Q}(\sqrt{577})}$, or have I just not looked far enough?
EDIT: Thanks to quid for editing my question to be more precise in regards to terminology. I am editing the question now because in my non-UFD example I used $24$ when I should have used $12$. It seems no one would have caught that but it was bothering me.
 A: Here is a theoretical approach, which uses a bit of class field theory to show that there will be infinitely many such primes.
Let $H$ be the Hilbert class field of $K=\mathbb Q(\sqrt{577})$, the maximal, everywhere unramified abelian extension of $K$.

Theorem: A prime ideal $\mathfrak p\subset\mathcal O_K$ is principal if and only if it splits completely in $H$.

You are looking for rational primes $p$ which split in in $\mathbb Q(\sqrt{577})$ and such that the prime ideals above $p$ are principal. Hence, the primes you are after (except maybe for the finitely many ramified primes) are exactly the primes which split completely in $H$. The Cebotarev density theorem guarantees that there will be infinitely many such primes.
This argument generalises to any number field $K$: either it is a UFD or there are infinitely many primes which are norms from $K$.

The following Sage code spits out some of the primes you're after ($577$ is missing because it is ramified)
K.<a> = NumberField(x^2 - 577)
L = K.hilbert_class_field('b')

for p in prime_range(15000):    
    splits = false
    for P in L.primes_above(p):
        if P.residue_class_degree() == 1 and P.absolute_ramification_index() == 1: 
            splits = true
            break
    if splits: print p

With output beginning
283 293 433 541 569 719 787 941 1097 1187 1429 1451 1531 1579 1663 1867 2003 2029 

A: The theoretical answer above is correct (although this answer has been edited to be less informative), but one can also give a concrete example.
$$(719)=(36+\sqrt{577})(36-\sqrt{577})$$
A: I honestly don't believe it, but I don't have the means to simulate a bunch of tries at the moment. Here is a good reason to believe you need to try harder though:  in order to be a norm $p$ must be a quadratic residue mod $577$. But this gives it about a 50/50 shot a priori. Combine that with the fact that this field has class number $7$ and if you assume (possibly not true, but heuristically) these two facts are independent you only get $1/14$ chance any given prime should be both a residue and a factor represents a principal ideal class. I don't know of any major theorems on the subject without more research, but I would try harder to find some if you're really vested in it, I would bet most anything it's just that you've picked a ring with a rather large class number.

The "blah" part of this is that this field has class number $7$ (in fact it is the smallest $d>0$ so that $\Bbb Q(\sqrt{d})$ has class number $7$.) If it were a UFD it would be easy, since you can use Dirichlet's theorem on infinitely many primes in arithmetic progressions to find $p\equiv 1\mod 577$. Then $p$ would be a norm in the integer ring by Quadratic Reciprocity, which of course means $p = a^2+ab-144b^2$ for some $a,b\in\Bbb Z$, so the ideal $\left(a+b{1+\sqrt{577}\over 2}\right)$ would have prime norm, hence be prime itself and manifestly principal.
Also note that, in theory you would just need $p\equiv a\mod 577$ with $a$ a quadratic residue, but since I don't know any off the top of my head for such a large modulus, I chose $1\mod 577$ which means the prime must be at least $2\cdot 577+1=1155$ to fit this prescription.
