Proving the Evenness of Fibonacci Numbers through Division by 3

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Here is the question:
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Prove that f_{n} is even if and only if n is divisible by 3. (f_{n} is of course the nth Fibonacci number)
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Proving that n is divisible by 3 => f_{n} is even is easily done by induction, but the other implication is eluding me. It is easy to show that f_{n} is even iff f_{n-3} 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 f_{n} is even iff f_{n-3} 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 f_0, f_1, and f_2.
 
I was thinking about that, and I think you are absolutely right. Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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