Prove by PMI $\gcd(f_n,f_{n+1}) = 1$ for all natural numbers $n$. $f_n$ represents the Fibonacci sequence.
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What are your own thoughts? – Arthur Oct 22 '14 at 12:48
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Three consecutive terms in a Fibonacci sequence can be written as $a,b,a+b$. If $d$ divides $b$ and $d$ divides $a+b$, then $d$ divides $a+b-b=a$.
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Hint: $\gcd(a,b)=\gcd(a-b,b)$.
Hagen von Eitzen
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@Ron.J.Adams By applying any of the definitions of $\gcd$ or one step of the Euclidean algorithm – Hagen von Eitzen Oct 22 '14 at 13:01
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