Two other formulas from part 16 contain terms with the factor F(m), and so can be treated similarly to the previous post. These are:
F(kn + m) ≡(−1)(n+1)k/2 F (m) (mod L(n)), for k even.
L(kn+m) ≡(−1)(n+1)(k−1)/2 F(m) L(n−1) (mod L(n)),for k odd.
Take m = 0 and use the fact that F(0) = 0 to see that:
L(n) |F (kn) for k even.
L(n) |L (kn) for k odd.
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