How do we use the hint to show the divisibility? I want to show that if $m=n^{13}-n$ and $n>1$ then $30290 \mid F_m$. (Hint: Show first that $a^{13} \equiv a \mod{2730}$.)
$F_m$ is the $m$-th Fibonacci number.
I have shown the hint as follows:
$2730=2 \cdot 3 \cdot 5 \cdot 7 \cdot 13$.
Using Fermat's little theorem, we deduce that $a^{13}\equiv a \pmod{5}$, $a^{13}\equiv a \pmod{2}$, $a^{13}\equiv a \pmod{3}$, $a^{13}\equiv a \pmod{7}$ and $a^{13}\equiv a \pmod{13}$.
Since $2,3,6,7,13$ are all relatively prime, we deduce that $2730 \mid a^{13}-a$.
But how can we use the fact that $a^{13} \equiv a \mod{2730}$ in order to deduce that $30290 \mid F_m$ ? 
 A: Let's look at the first few Fibonacci numbers. In particular $F_2$, $F_3$, $F_5$, $F_7$, and $F_{13}$. The sequence is 
$$ 1,\mathbf{1},\mathbf{2},3,\mathbf{5},8,\mathbf{13},21,34,55,89,144,\mathbf{233}.$$
We can ignore $F_2=1$, but $F_3=2$, $F_5=5$, $F_7=13$, and $F_{13}=233$ are all distinct primes, and their lcm is therefore the product $2\cdot 5\cdot 13\cdot 233=30290$.
Finally we link this fact with the part that you have shown by noting that if $n\mid m$, then $F_n\mid F_m$.
Proof of the fact:
The Fibonacci sequence is also given by the following formula:
$$\newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}}
\bmat F_{n+1}\\F_n\emat=\bmat 1&1\\1&0 \emat^n\bmat 1\\ 0\emat.$$
Now suppose for contradiction that $F_n\nmid F_m$ for some integers $n$ and $m$ with $n\mid m$. Then let $k$ be the least positive integer such that $F_n\nmid F_{kn}$. Then we have that
$$\bmat F_{kn+1}\\F_{kn}\emat
=\bmat 1&1\\1&0 \emat^{kn}\bmat 1\\ 0\emat
=\bmat 1&1\\1&0 \emat^{(k-1)n}\bmat 1&1\\1&0 \emat^{n}\bmat 1\\ 0\emat
=\bmat 1&1\\1&0 \emat^{(k-1)n}\bmat F_{n+1} \\ F_n \emat.
$$
Taking this equation mod $F_n$, we see that
$$\bmat F_{kn+1}\\F_{kn}\emat
=\bmat 1&1\\1&0 \emat^{(k-1)n}\bmat F_{n+1} \\ 0 \emat
=F_{n+1}
\bmat 1&1\\1&0 \emat^{(k-1)n}\bmat 1 \\ 0 \emat
=F_{n+1}
\bmat F_{(k-1)n+1} \\ F_{(k-1)n} \emat,
$$
so mod $F_n$, we have $F_{kn}=F_{n+1}F_{(k-1)n}$, but by assumption $k$ was the least positive integer such that $F_n\nmid F_{kn}$, so $F_n\mid F_{(k-1)n}$ contradiction.
A: $F_i$ obey $\, j\mid k\,\color{#c00}{\overset{\rm D}\Rightarrow}\, F_j\mid F_k,\,$ and $\,\gcd(F_j,F_k) = F_{\gcd(j,k)},\,$ being a strong divisibility sequence 
By here $\ 3,\ 5,\ 7,\ 13\ \mid n^{13}-n\ $ for all integers $n,\,  $ by $\, \ 2,4,6,12\mid 13\!-\!1$
$\ \, \color{#c00}{\overset{\rm D}\Rightarrow}\ F_3,F_5,F_7,F_{13}\mid F_{\large n^{13}-n}\ $ for all integers $n$
So their lcm = product $= 2\cdot 5\cdot 13\cdot 233 = 30290$ also divides   $F_{\large n^{13}-n}\,$ since $F_i$ with coprime indices are coprime by the above gcd formula. $ $ QED
