I was reading a book, where the author said, with no proof, that
No four distinct fibonacci numbers are in arithmetic progression.
I tried with no progress for about one hour.
Can anyone show me a proof, please?
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1No proof in that link too. – Muhammad Rasel Parvej Dec 22 '16 at 05:58
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Interesting link @lab bhattacharjee but unfortunately the proof is missing – marwalix Dec 22 '16 at 05:59
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I have found a proof. What should I do now? – Muhammad Rasel Parvej Dec 22 '16 at 06:10
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1@MuhammadRaselParvej I would be interested in seeing the proof; I believe you can answer your own question? – TomGrubb Dec 22 '16 at 06:11
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Yes answer your own question – marwalix Dec 22 '16 at 06:13
1 Answers
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Let's assume $F_h < F_i < F_j < F_k$ are in AP, so $F_i — F_h = F_k — F_j=d$, the common difference
Then $d = F_i-F_h < F_i$; on the other hand, $d = F_k-F_j \geq F_k-F_{k-1}=F_{k-2}\geq F_i$, which poses a contradiction.
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1$F_{k-2}\geq F_i$ implies $d\geq F_i$. Isn't it a contradiction to $d<F_i$? – Muhammad Rasel Parvej Dec 22 '16 at 06:45
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1Why would $F_k-F_j \geq F_k-F_{k-1}$ imply that there is an extra fibonacci number between $F_k$ and $F_{k-1}$? – Muhammad Rasel Parvej Dec 22 '16 at 06:48
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2No, it's okay. Minus changes the direction of the inequality. – Muhammad Rasel Parvej Dec 22 '16 at 06:59