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Start with the Fibonacci numbers, F(n), dividing them successively by each natural number, N, in turn. At some point N will divide F(n), For example, 2 divides itself, which is F(3), and 3 divides itself, which is F(4). These are what OEIS calls the Fibonacci entry points, Fibonacci indexes {1, 3, 4, 6, 5, 12,...}, OEIS A001177.
Now each N is periodic with respect to the Fibonacci series so must eventually divide some later Fibonacci number, index F(n+m) as well.
The full 'Pisano period' of each N is given in OEIS A001175 {1, 3, 8, 6, 20, 24, 16,...}. Frequently, a full Pisano period will contain more than one case where N divides some F(n). Fortunately, the 'zeros', i.e. the cases where N divides successive F(n), are evenly distributed along the period.
(These zeros are given in OEIS as A001176 {1, 1, 2, 1, 4, 2, 2,...}; there are apparently always either 1, 2, 4, or 6 occasions on which this occurs. For example, 2 divides 8, which is F(6), at the end of its period but 3 divides 21, which is F(8}, and 144, which is F(12), while 5 divides 55, 610, and 1353, which are F(10), F(15) and F(20) respectively.)
Fortunately, for each N we need only consider the Fibonacci entry point, F(n) in the problem, and the first available zero, F(n+m), after that.
Remember that the zeros are distributed evenly along the period. But the entire Fibonacci sequence also starts with a zero; F(0) = 0, so there must be the same number of indices between 0 and F(n) as there are between F(n) and F(e+m); i.e. n=m.
However, we know from Binet that F(2n) = F(n)L(n), explaining the relation, so it is always true that F(n+m) /F(n) = L(m). The weaker result follows automatically.
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