# 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.

• It's certainly laborious to compute $5^{44} \pmod{89}$ by hand, and a mistake is almost certain to creep in along the way. However, thanks to quadratic reciprocity, you know the Legendre symbols $$\left(\frac{5}{89}\right) = \left(\frac{89}{5}\right)$$ and you can just do $89^2 \equiv 1 \pmod 5$. – Robert Soupe May 21 '17 at 1:34
• If $F_n$ is prime, then $n$ is prime. In this case, $F_n$ remains prime in the field $\mathbf{Q}(\sqrt{5})$ if $n \equiv 3,7 \mod 10$, and splits if $n \equiv 1,9 \mod 10$. So, for example, $F_{3} = 2$, $F_{13} = 233$, $F_{23} = 28657$, $F_{43} = 433494437$, remain prime, $F_{7} = 13$, $F_{17} = 1597$, $F_{47} = 2971215073$ remain prime, whereas $F_{11} = 89$, $F_{131} = 1066340417491710595814572169$ split, as do $F_{29} = 514229$ and $F_{359}$ which equals $$475420437734698220747368027166749382927701417016557193662268716376935476241.$$ Proof: $F^2_{n}$ modulo $5$ is periodic modulo $10$. – Gordon May 21 '17 at 7:45

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

• $13^3 - 5 \cdot 4^4 = 89$ and $u^2 - 5 v^2$ represents exactly the same integers as $x^2 - xy - y^2.$ Any prime $p \equiv \pm 1 \pmod 5,$ or a product of such – Will Jagy May 20 '17 at 21:59
• Is there a name for this criterion? That's very cool. – law-of-fives May 20 '17 at 22:03
• @law-of-fives I think it's called the "Dedekind factorization criterion" or something like that. See here for the statement: math.stackexchange.com/questions/38431/… – Kenny Wong May 20 '17 at 22:06

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.

• Actually, you can go easier, modulo 5, since $1^2 \equiv 4^2 \equiv 1 \pmod 5$ and $2^2 \equiv 3^2 \equiv -1 \pmod 5$. – Robert Soupe May 26 '17 at 15:58
• We can simplify more. If $F_n$ is prime, then $n$ is odd (as $F_{2n} = F_n L_n$ where $L_n$ is a Lucas number). For $n$ odd, we have $F_n = \pm 1 \bmod 5$ if and only if $n = \pm 1 \bmod 5$. – David E Speyer Jul 18 '17 at 13:33

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

• Would it be possible for you to explain the method in more detail? I stared at the sums but couldn't quite figure out how to use them to factorise the principal ideals. Since I'm not a specialist algebraic number theorist, I don't know all the tricks of the trade, so this is most likely my fault! – Kenny Wong May 20 '17 at 22:17
• @KennyWong, the norm is the binary quadratic form $x^2 - xy - y^2,$ it might be a plus sign in the middle rather than the - I put. This form integrally represents exactly the same integers as $u^2 - 5 v^2.$ A number can only be primitively represented when all factors are either $5$ or primes $q \equiv \pm 1 \pmod 5.$ (indeed, primitively or not, for a number to be represented, all prime factors $q \equiv \pm 2 \pmod 5$ must occur to even exponents). From my little computer run, few Fibonacci numbers can be so expressed. – Will Jagy May 20 '17 at 22:24
• Thanks! I suppose the thing I'm missing is the link between integral/primitive representations by quadratic forms and factorisation of principal ideals. Do you know a web link that explains this at an elementary level? Of course, this could well be beyond me. :) – Kenny Wong May 20 '17 at 22:54
• @KennyWong how would you calculate the norm of the element $x + y \phi$ where $x,y$ are rational integers? I think it comes out $x^2 +xy - y^2.$ en.wikipedia.org/wiki/… – Will Jagy May 20 '17 at 23:05
• Oh I totally get it!!! $\mathbb Q(\sqrt 5)$ has class number one, so the prime factor I'm looking for is itself a principal ideal $\langle x + \phi y \rangle$ with $N(x + \phi y) = p$. (And yes, it comes out as $x^2 + xy - y^2$ with the plus sign rather than the minus sign.) Thank you very much for being so patient while I was being so dumb! :) – Kenny Wong May 20 '17 at 23:20