Define the Fibonacci numbers by $F_0=1$, $F_1=1$, and for each $n\ge2$, $F_n=F_{n-1}+F_{n-2}$. I need to prove that no Fibonacci number is twice that of another Fibonacci number (except for $F_2=2=2\times 1=2F_1$). I tried to use induction but I can't seem to figure it out. Thanks!
Let $n\in\mathbb{N}$. Show that $F_n \ne 2F_m$ for any $m\ge3.$
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combinatorics
fibonacci-numbers
1 Answers
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Sequence of Fibonacci numbers is, except for first two values, increasing. From it follows that $F_{n+1}=F_n+F_{n-1} > 2F_{n-1}$. So given value exceeds twice its value very quickly, only way to get $F_n=2F_m$ would be for two consecutive values. But that means $F_{n+1}=2F_n=F_n+F_{n-1}$ which implies $F_{n-1}=F_{n}$ which is true only for $n=1$.