Find all positive integers $m>1$ such that any sum of $m$ consecutive Fibonacci numbers is divisible by $m$.

Question

Let $\{u_n\} -$ Fibonacci sequence: $u_1=u_2=1, u_{n+1}=u_n+u_{n-1}, n\ge2$. Find all positive integers $m>1$ such that any sum of $m$ consecutive Fibonacci numbers is divisible by $m$.

My work so far:

1) Let $m=2$. Since $\gcd(u_n,u_{n+1})=1$, then $2 \not| u_n+u_{n+1}$

Hence $m\not =2$

2) Let $m=3$. Let $$u_n \equiv r_n (\bmod3)$$ $$\{r_n\}:\{1,1,2,0,2,0,2,..\}$$ hence, for $n>3$ $$u_n+u_{n+1}+u_{n+2}\equiv0+2+0=2 (\bmod3)$$ or $$u_n+u_{n+1}+u_{n+2}\equiv2+0+2\equiv1 (\bmod3)$$ Hence $m\not =3$

Similarly, $m\not =4$, $m\not= 5$

Addition

I used Pisano period. There $m=\Pi(m)=24$.

\begin{bmatrix} r_n & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 12 \\ r_n & 10 & 7 & 17 & 0 & 17 & 17 & 10 & 3 \\ r_n & 13 & 16 & 5 & 21 & 2 & 23 & 1 & 0 \end{bmatrix}

Then $$u_n+u_{n+1}+...+u_{n+23}\equiv r_n+r_{n+1}+...+r_{n+23}=216=24\cdot9 \equiv0(\bmod 24)$$ Hence, $m=24$

Similarly, $m =48$, $m=72$

Is a finite or infinite the set of such numbers $m$?

Answer 1

We need the result that the sum of $m$ consecutive Fibonacci numbers starting with $F_k$ is $$F_{m+k+1}-F_{k+1}$$ See Successively longer sums of consecutive Fibonacci numbers: pattern?

But since $F_{m+k+1}-F_{k+1}=(F_{m+k}-F_k)+(F_{m+k-1}-F_{k-1})$ it is necessary and sufficient for the question that $m$ divides $F_m-F_0=F_m$ and $F_{m+1}-F_1=F_{m+1}-1$.

The sequence of $m$ such that $m|F_m$ is well known http://oeis.org/A023172 but unfortunately not that well characterised. Every term greater than 1 is divisible by 5 or 12, but not all such numbers are in the sequence.

Mod 12 the sum of the first $k$ terms of the Fibonacci sequence has period 24 (you can easily evaluate 1,2,4,7,0,8, ...) and is 0 for $k=0,15,21,23\bmod24$. But 12 does not divide $24h+15,24h+21$, $24h+23$, so that means that for $m$ to divide the sum of consecutive terms we need a multiple of 12 to be a multiple of 24.

Similarly, mod 5 the sum of the first $k$ terms mod 5 has period 20 and is 0 for $k=6,17,19,20\bmod20$. So multiples of 5 could only work if they were also multiples of 20 and hence also multiples of 4.

Mod 4 the sum of the first $k$ terms of the Fibonacci sequence has period 6: 1, 2, 0, 3, 0, 0, and is 0 for $k=3,5,6\bmod6$, but $6h+3,6h+5$ are not multiples of 4, so multiples of 4 only work if they are also multiples of 6 and hence of 12, and hence of 24. So multiples of 5 do not work (for the sum of consecutive terms) unless they are also multiples of 24.

Thus a necessary condition is that $m$ is a multiple of 24. But unfortunately that is not a sufficient condition. For example, 168 does not work. Nor does 264, 312, 408, 456, 504 and no doubt many others.

---------- Added later ----------

Mod 7 the sum of the first $k$ terms is 0 for $k=0,4,13,15\bmod16$. So the sum of the first $k$ terms is only divisible by 7 if $k$ is an even multiple of 24. That rules out odd multiples of 168 (ie 168, 504, 840, ...).

Similarly, mod 11 the sum of the first $k$ terms is 0 for $k=0,7,9\bmod10$. So the sum of the first $k$ terms is only divisible by 11 if $k$ is also divisible by 10. So multiples of $11\cdot24=264$ must also be multiples of 5. That rules out $264,528,792,\dots$.

Similarly, mod 13 the sum of the first $k$ terms is 0 for $k=0,10,25,27\bmod28$. So the sum of the first $k$ terms is only divisible by 13 if it is also divisible by 28 and hence by 7 and hence by 16. So that rules out $13\cdot24=312h$ unless $h$ is a multiple of $14$. So we rule out $312,2\cdot312,3\cdot312,\dots,13\cdot312,15\cdot312\dots$.

Similarly, mod 17 the sum of the first $k$ terms is 0 for $k=0,14,33,35\bmod36$. So the sum of the first $16k$ terms is only divisible by 17 if it is also divisible by 9. In other words $17\cdot24h=408h$ is ruled out unless $h$ is a multiple of $3$. So we rule out 408,816 etc.

Summarising, $1\cdot24,2\cdot24,3\cdot24,4\cdot24$, $5\cdot24,6\cdot24,8\cdot24,9\cdot24$, $10\cdot24,12\cdot24,14\cdot24$, $15\cdot24,16\cdot24$ all work. But we have established infinitely many multiples of 24 that do not work and only finitely many that do work. One could continue the laborious process of examining primes, but a full solution needs some new ideas.