Aside from $5$, are all prime Fibonacci numbers also prime in $\mathbb{Z}[\phi]$?

Solution 1:

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 $$ \langle 89 \rangle = \langle 89, \phi - 10 \rangle \langle89 , \phi - 80\rangle.$$

Solution 2:

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.

Solution 3:

next $$ 10 - 5 \cdot 3^2 = 55 = 5 \cdot 11$$ $$ 13 - 5 \cdot 4^2 = 89$$ $$ 134 - 5 \cdot 7^2 = 17711 = 89 \cdot 199$$ $$ 305 - 5 \cdot 65^2 = 75025 = 5^2 \cdot 3001$$ $$ 893 - 5 \cdot 238^2 = 514229$$ $$ 26129 - 5 \cdot 10170^2 = 165580141 = 2789 \cdot 59369 $$

is the next prime Fibonacci that is $\pm 1 \pmod 5$ therefore can be written as $x^2 + xy - y^2$ in integers. $$ \color{blue}{ (u+v)^2 + (u+v)(-2v)- (-2v)^2 = u^2 - 5 v^2} $$

8 =  2^3
13 =  13
21 =  3 7
34 =  2 17
55 =  5 11
89 =  89
144 =  2^4 3^2
233 =  233
377 =  13 29
610 =  2 5 61
987 =  3 7 47
1597 =  1597
2584 =  2^3 17 19
4181 =  37 113
6765 =  3 5 11 41
10946 =  2 13 421
17711 =  89 199
28657 =  28657
46368 =  2^5 3^2 7 23
75025 =  5^2 3001
121393 =  233 521
196418 =  2 17 53 109
317811 =  3 13 29 281
514229 =  514229
832040 =  2^3 5 11 31 61
1346269 =  557 2417
2178309 =  3 7 47 2207
3524578 =  2 89 19801
5702887 =  1597 3571
9227465 =  5 13 141961
14930352 =  2^4 3^3 17 19 107
24157817 =  73 149 2221
39088169 =  37 113 9349
63245986 =  2 233 135721
102334155 =  3 5 7 11 41 2161
165580141 =  2789 59369
267914296 =  2^3 13 29 211 421
433494437 =  433494437
701408733 =  3 43 89 199 307
1134903170 =  2 5 17 61 109441