Proving the Evenness of Fibonacci Numbers through Division by 3

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SUMMARY

The discussion focuses on proving that the nth Fibonacci number, denoted as f_{n}, is even if and only if n is divisible by 3. The proof for the implication that if n is divisible by 3 then f_{n} is even is established through mathematical induction. The challenge lies in proving the converse, that if f_{n} is even, then n must be divisible by 3. Participants suggest leveraging the property that f_{n} is even if and only if f_{n-3} is even, indicating a recursive approach to the proof.

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mattmns
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Here is the question:
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Prove that [itex]f_{n}[/itex] is even if and only if n is divisible by 3. ([itex]f_{n}[/itex] is of course the nth Fibonacci number)
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Proving that n is divisible by 3 => [itex]f_{n}[/itex] is even is easily done by induction, but the other implication is eluding me. It is easy to show that [itex]f_{n}[/itex] is even iff [itex]f_{n-3}[/itex] is even, but I can't see if this helps. Any ideas about how to prove this implication? Thanks.
 
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mattmns said:
It is easy to show that [itex]f_{n}[/itex] is even iff [itex]f_{n-3}[/itex] is even, but I can't see if this helps.
If this is easy to show, then it would be enough to look at the evenness of [tex]f_0, f_1[/tex], and [tex]f_2[/tex].
 
I was thinking about that, and I think you are absolutely right. Thanks.
 

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