Solution:

The elements of a number bracelet $N$ will be denoted $N_{0}, N_{1}, N_{2}, \ldots$; $N_{0}$ and $N_{1}$ being the starting numbers. Let $F$ be the bracelet with starting numbers $F_{0}=0, F_{1}=1$. All congruences are modulo $m$ unless otherwise noted.

Since there are only $m^{2}$ possible pairs $(N_{k}, N_{k+1})$, there must be repetition, so that
$$(N_{k}, N_{k+1}) = (N_{p+k}, N_{p+k+1})$$
for some $p \geq 1$ and $k$. If this is true for one $k$, it must be true for the next $k$, since $N_{k+1} = N_{p+k+1}$ by hypothesis and $N_{k+2} \equiv N_{k} + N_{k+1} = N_{p+k} + N_{p+k+1} \equiv N_{p+k+2}$, and so it is true for all $k$ bigger than that $k$. It must also be true for the previous $k$, since $N_{k} = N_{p+k}$ by hypothesis and $N_{k-1} \equiv N_{k+1} - N_{k} = N_{p+k+1} - N_{p+k} \equiv N_{p+k-1}$, and so it is true for all $k$ smaller than that $k$. Therefore it is true for all $k$.

Let $P_{N}$, the period of $N$, be the smallest $p$ such that this is true for some $k$ (and therefore for all $k$).

It is clear that if $0 \leq i, j < P(N), i \neq j$, then $(N_{i}, N_{i+1}) \neq (N_{j}, N_{j+1})$. Therefore, among the terms $N_{0}$ through $N_{P_{N}}$, there is no repetition even of pairs of adjacent terms. But because of the property of $P_{N}$, $(N_{0}, N_{1}) = (N_{P_{N}}, N_{P_{N}+1})$ and the repetition will continue. Thus the number bracelet $N$ consists of a single loop of $P_{N}$ elements.

We will prove by induction that for all $n \geq 1$,
$$N_{n} \equiv N_{0} F_{n-1} + N_{1} F_{n}.$$
The case $n=1$ is trivial, and $n=2$ follows directly from the definition of $N$. If this is true for $n=k$ and $n=k-1$, $N_{k+1} \equiv N_{k} + N_{k-1} \equiv N_{0} F_{k-1} + N_{1} F_{k} + N_{0} F_{k-2} + N_{1} F_{k-1} = N_{0} F_{k} + N_{1} F_{k+1}$ so it is also true for $n=k+1$.

It is clear from the circularity of the number bracelet that $N_{k+R} = N_{k}$ for all $k \geq 1$ if and only if $P_{N} \mid R$. Since $N_{k+P_{F}} = N_{0} F_{P_{F}+k-1} + N_{1} F_{P_{F}+k} = N_{0} F_{k-1} + N_{1} F_{k} = N_{k}$ for all $k \geq 1$, $P_{N} \mid P_{F}$. This completes the proof.