= Solution
The base case is $F_5=5$, so $5\mid F_5$. To relate successive blocks of five, use the <Fibonacci addition formula>:
$$
F_{k+5}=F_5F_{k+1}+F_4F_k=5F_{k+1}+3F_k.
$$
This identity can also be obtained simply by applying the <Fibonacci recurrence> five times. If the <mathematical induction> hypothesis is $5\mid F_{5n}$, then
$$
F_{5(n+1)}=5F_{5n+1}+3F_{5n}
$$
is a sum of two multiples of five. Hence
$$
\boxed{F_5\mid F_{5n}\quad\text{for every integer }n>0.}
$$
The generalisation is <Fibonacci multiple-index divisibility>:
$$
\boxed{F_m\mid F_{mn}\quad(m,n\geq1).}
$$
Indeed, $F_{mn+m}=F_mF_{mn+1}+F_{m-1}F_{mn}$, so precisely the same <mathematical induction> works with $5$ replaced by $m$.
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