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Let $f_n$ be Fibonacci Sequence. $$gcd(f_{n},f_{n+2})=1,\quad \forall\,n\in\mathbb{N}.$$ Prove

Could you help me with this one? I have done the base case, I just can't figure out the inductive step. Thanks!

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2 Answers 2

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Suppose that for a particular $k$ we have $\gcd(f_k,f_{k+2})=1$. We will show that $\gcd(f_{k+1},f_{k+3})=1$.

Suppose that $d$ divides $f_{k+1}$ and $f_{k+3}$. Then since $f_{k+2}=f_{k+3}-f_{k+1}$, we have $d$ divides $f_{k+2}$. But then because $f_{k}=f_{k+2}-f_{k+1}$, it follows that $d$ divides $f_k$.

This forces $d=1$ by the induction assumption.

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A broad hint: $\gcd(f_{n+3}, f_{n+1})$ $=\gcd(f_{n+2}+f_{n+1}, f_{n+1})$ $=\gcd(f_{n+2}+f_{n+1}-f_{n+1}, f_{n+1})$ $=\gcd(f_{n+2}, f_{n+1})$ $=\gcd(f_{n+2}, f_{n+1}-f_{n+2})$ $=\ldots$, where the second and last steps use the theorem that $\gcd(a+cb, b) = \gcd(a,b)$.

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