Fibonacci

F_24k+12 is divisible by 144 but not by 288.

We make use of the formula:
F_pL_q + L_pF_q = 2F_p+q ( see 15.) )

We proof (by induction) that F_24k+12 is divisable by 144 but not by 288.
The statement is true for k=0, for F₁₂ = 144 is divisable by 144 and not by 288.
Now, assume that the statement is correct for k=n. Then:
2F_24(n+1)+12 = F_24n+12L₂₄ + L_24n+12F₂₄
F_24n+12 = 144*(2u+1) for certain u (according to the hypothesis with k=n).
L₂₄ = 103682 = 2*51841.
L_24n+12 = 2*(2v+1) for certain v (see below the line).
F₂₄ = 46368 = 2*144*161.
Hence 2F_24(n+1)+12 = 144*(2u+1)*2*51841 + 2*(2v+1)*2*144*161 or:
F_24(n+1)+12 = 144*((2u+1)*51841 + 2*(2v+1)161).
This number can be divided by 144, but not by 288.
This shows that the statement is also correct for k=n+1.
Apparently, F_24k+12 is divisable by 144 but not by 288, for every k.



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Statement: For every k is L_6k even but not divisable by 4.
With induction.
L₀ = 2, thus the statement is correct for k=0.
Assume n≥0 and L_6n = 2(2u+1) for certain u.
We will show that L_6(n+1) = 2(2v+1) for certain v.
Formulae 16.) says
L_pL_q + 5F_pF_q = 2L_p+q
So 2L_6(n+1) = L_6nL₆ + 5F_6nF₆ =
2(2u+1).18 + 5*F_6n*8.
Hence L_6(n+1) = 2*(9*(2u+1) + 2*L_6n) = 2*(2v+1) for certain v.
Therefore, the statement is correct for k=n+1 too, and consequently for every k.