Fibonacci divisibility

Question

Is $2051$ a factor of any fibonacci number?

It is not a factor of any perfect number. The prime factors of $2051$ are $7$ and $293$, which are both prime. the $8$th fibonacci number, is the first time that $7$ appears as a factor, and the $147$th fibonacci is the first time that $293$ appears. $293$ also appears as a factor at the $294$th fibonacci. I don't know if either appear at the $301$st fibonacci. ($300$ being the sum of the factors). My hypothesis is that $2051$ and it's multiples are a constant to link fibonacci and the perfect numbers- $1,2,4,8,16$.

(GP) ? for(i=1,10000,if(fibonacci(i)%2051==0,print(i);break(),)) -> 1176, and indeed 1176th fibonacci number is divisible by 2051. — JiminP, Dec 02 '13 at 06:24
Has it been proved that $2051$ is not a factor of an odd perfect number? Oh my bad, it appears that daniel refers to known almost perfect numbers when he says perfect number. — Jose Arnaldo Bebita Dris, Jul 11 '16 at 11:01

John Dvorak · Answer 1 · 2013-12-04T08:31:02.250

Yes.

The Fibonacci sequence is periodic modulo any base. This should be clear from the definition $a_n = a_{n-1} + a_{n-2}$ - if any two consecutive values repeat, then the whole sequence repeats as well - and from the pigeonhole principle - there are only so many possible pairs.

Thus, a zero must appear in the sequence. The first two terms are $1, 1$. Find the next occurence of $1, 1$. The previous term in the sequence is a zero.

Any value that is zero modulo $n$ is divisible by n for every positive integer $n$. Also, Fibonacci modulo-n contains a zero for every positive integer $n$.

Thus, for every positive integer $n$, there is a term in the Fibonacci sequence that is divisible by $n$.

this can be generalised to arbitrary integer sequences where

each term can defined recursively from $k$ previous terms using only operations that "work modulo".
the same applies backwards - each term can be defined from $k$ following terms using only operations that "work modulo".
the sequence can be extended backwards to include zero

These operations includes addition, subtraction and multiplication, but it excludes division where the divisor and the base are not coprime.

Namely:

Any sequence defined by a linear recurrence relation with integer coefficients is eventually periodic modulo every base with the period not greater than the square of the base.

Any such sequence is periodic modulo every base that is coprime with the leading term of the recurrence. Namely, it is periodic modulo every base if the leading term is $1$.

For any such sequence that can be extended backwards to include a zero, for each positive integer $n$ coprime with the leading term, there are infinitely many terms in the sequence divisible by $n$, and the distance between any two such consecutive terms is never greater than $n^2$.

There's a subtlety here that might deserve explicit mention. The pigeonhole argument shows that the sequence is eventually periodic. Since you want full periodicity, you are also using the fact that the recurrence is reversible: $a_{n-2} = a_n - a_{n-1}$ is uniquely determined by the two terms after it without any ambiguity caused by zero divisors mod $n$ (this is not necessarily the case for general linear recurrences with integer coefficients). — Erick Wong, Dec 02 '13 at 07:47
@ErickWong there is another unstated assumption that is required when generalising. Generalised. Do you find said subtlety as stated explicitly enough after the edit? — John Dvorak, Dec 02 '13 at 10:23
Yes, looks quite good to me, +1. One tiny quibble is I'm not sure in what sense exponentiation can be said to "work modulo". Exponentiation by fixed powers perhaps, but this doesn't necessarily apply backwards (yes in the case of $x_{n+1} = x_n^2 + x_{n-1}$, but not in the case of $x_{n+1} = x_n^2$). — Erick Wong, Dec 04 '13 at 04:59
@ErickWong you are right, sorry; exponentiation only works in special cases; removed — John Dvorak, Dec 04 '13 at 08:31

score 2 · Answer 2 · edited Apr 13 '17 at 12:20

2

From this answer you can see that $7|F_n$ iff $8|n$.

By a similar method, you note that $293|F_{147}$ and $293|F_{294}$, so it will appear when $147|n$.

Clearly, then, we want to consider $8\cdot147 = 1176$.

In particular, re-referencing the linked answer, we have that $7\cdot293 = 2051|F_{1176}$, according with the comment above.

edited Apr 13 '17 at 12:20

Community

1

answered Dec 02 '13 at 06:31

Benjamin Dickman

14,451

score 1 · Answer 3 · answered Dec 02 '13 at 06:31

I am not sure what your hypothesis on perfect numbers is, but consider this:

Any two adjacent residues mod $n$ in a Fibonacci sequence, determine the entire sequence mod $n$.

Also, there are only $n^2$ possible pairs of residues so you know it has to repeat.

So you can trace the Fibonacci sequence mod 7; there will be a cycle length $k$, so that every $k$th Fibonacci number ends up as $0 \text{(mod 7)}$. ($k$ = 8)

Same thing for 293. And gcd(147, 8) = 1176 so you know the 1176th Fibonacci number is divisible by $293 * 7 = 2051$.

Our friend Wolfram Alpha tells us this number is 263014645338081959394002434587551312768847375888603586237774998502936063023131672433074857898670115025850753711767851493503868182178181357041212318303125912580537357570323167455972378184345773715225354264871395045042309350710402468383175624928032, which is indeed divisible by 2051.

Fibonacci divisibility

3 Answers3