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This is the question: The Fibonacci numbers are recursively dened by $f_1 = f_2 = 1$, and $f_n = f_{n-1} + f_{n-2}$ for $n > 1$. Prove that every fourth Fibonacci number is a multiple of $3$.

I've been stuck on this question for a while. If someone could help me out or at least help me start it, that would be great

Maadhav
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SCOW9843
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    Duplicates: https://math.stackexchange.com/questions/815004/determine-which-fibonacci-numbers-are-even, https://math.stackexchange.com/questions/972693/prove-that-for-each-fibonacci-number-f-4n-is-a-multiple-of-3, https://math.stackexchange.com/questions/60340/fibonacci-modular-results-and-gcdf-n-f-m-f-gcdn-n – Sam OT Nov 02 '17 at 22:06
  • Hint: Try proceeding by induction. – Daniel Robert-Nicoud Nov 02 '17 at 22:01

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You can then show that $$ f_{4(n+1)} = 2 f_{4n} + 3 f_{n+1} $$ by just repeatedly plugging in the formula. As Daniel suggests, use this along with induction, and you should hopefully be able to get the result! :)

Originally I was just going to write "start with $f_{4(n+1)}$ and repeatedly plug in the formula". Since seeing this answer to the same question elsewhere, I added the right-hand side. Wasn't really anything, but felt I should credit it all the same!

Sam OT
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