Aside from $5$, are all prime Fibonacci numbers also prime in $\mathbb{Z}[\phi]$? Given $$\phi = \frac{1 + \sqrt 5}{2},$$ it follows that in $\mathbb{Z}[\phi]$ we have $5 = (\sqrt 5)^2 = (-1 + 2 \phi)^2$, as it is obvious that $\langle 5 \rangle$ is a ramifying ideal.
But what about other Fibonacci numbers that are also prime? $13$ is also prime, I have triple-checked it. It looks like $89$ is also prime but I feel like I have made a mistake somewhere along the way. I'll start by doublechecking my Jacobi symbol calculation.
Clearly there are primes that split in $\mathbb{Z}[\phi]$. But when those primes are also Fibonacci numbers, is their primality also guaranteed in $\mathbb{Z}[\phi]$?
EDIT: I did make a mistake in my Legendre symbol calculation. That prevented me from finding $$\left(\frac{7}{2} - \frac{9 \sqrt 5}{2}\right) \left(\frac{7}{2} + \frac{9 \sqrt 5}{2}\right) = (8 - 9 \phi)(-1 + 9 \phi) = -89.$$ I am grateful for all the answers, whether already posted or yet to post.
 A: I want to believe that there is some deep connection between Binet's formula for the Fibonacci numbers and splitting of primes in $\textbf Z[\phi]$. Looking at the formula discourages me from this line of thought, however:
$$F_n = \frac{\phi^n - (1 - \phi)^n}{-1 + 2 \phi}$$
At their core, the Fibonacci numbers seem to be essentially additive in nature, rather than multiplicative. Note also that $N(\phi) = N(1 - \phi) = -1$.
What really matters for determining if prime $F_n$ splits or is inert in $\textbf Z[\phi]$ is its congruence modulo $20$. We know that from $13$ on, prime $p = F_n \equiv 1 \pmod 4$. Then, by quadratic reciprocity, $$\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right) = p^2 \bmod 5.$$ If $p \equiv 1 \textrm { or } 9 \bmod 20$, then $p^2 \equiv 1 \bmod 5$. But if $p \equiv 13 \textrm { or } 17 \bmod 20$, then $p^2 \equiv -1 \bmod 5$.
Therefore, $13, 233, 1597, 28657, 433494437, 2971215073, \ldots$ are inert in $\textbf Z[\phi]$, while $89, 514229, 1066340417491710595814572169, \ldots$ split.
This is not to say that there isn't a deep connection between Binet's formula and $\textbf Z[\phi]$. It might exist but it might be much too deep for me to see it.
A: If I've done my sums correctly, $89$ is not prime in $\mathbb Z[\phi]$.
The minimal polynomial of $\phi$ is $t^2 - t - 1$, which factorises as
$$ t^2 - t - 1 = (t - 10)(t - 80)$$
in the ring $\mathbb Z_{89}[t]$.
Therefore, using a criterion by Dekekind, we find that the principal ideal $\langle 89 \rangle $ factorises as
$$ \langle 89 \rangle = \langle 89, \phi - 10 \rangle \langle89 , \phi - 80\rangle.$$
