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Prove that Fibonacci sequence contains An infinite Sub-sequence That all its element are co-prime.

Prove Two Sub-sequence different

Instruction: Fermat numbers.

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attempt 1: $F_n$ is strong divisibility sequence , $\gcd(F_m,F_n)$=$F_{\gcd(m,n)}$ we need to prove there an infinite sequence of coprime integers.

ATB
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1 Answers1

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Hint: by the simple proof in this answer, the Fibonacci sequence $\,F_n\,$ is strong divisibility sequence, i.e. $\,\gcd(F_m,F_n) = F_{\gcd(m,n)},\,$ so the problem reduces to proving that there is an infinite sequence of coprime integers (indices), for which there are many obvious methods (some that don't require primes, e.g. Euclid's famous constructive algorithm that recursively generates an infinite sequence of (co)prime integers).

Bill Dubuque
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  • Dubque Thanks for the Hint, i read all the link's u sent. i have already prove that gcd(Fm,Fn)=Fgcd(m,n) but i don't get it how i can prove that have an infinite sequence of coprime integers – ATB Nov 17 '20 at 17:18
  • @atb The final link above shows how Euclid's method constructs an infinite sequence of pairwise coprime integers (or primes). Use any such sequences as indices for $F$ to map it to pairwise coprime fib's. – Bill Dubuque Nov 18 '20 at 00:44
  • Thanks, understood but how its indices for F to map it to pairwise coprime fib's i don't get it . sorry. – ATB Nov 18 '20 at 12:41
  • @atb If $,i\neq j\Rightarrow, \color{#c00}{\gcd(n_i,n_j)=1},$ then $,\gcd(F_{n_i},F_{n_j}) = F_{\color{#c00}{\gcd(n_1,n_j)}}! = F_{\color{#c00}{1} }=1\ \ $ – Bill Dubuque Nov 18 '20 at 16:39
  • I understand what u write but how its can be An infinite Sub-sequence That all its element are co-prime in Fibonacci, you said its simple proof but I don't see it so easy, you maybe can rescue me and give a full proof I will apricate that . I read everything said in your post – ATB Nov 18 '20 at 18:36
  • @atb It's not clear what you don't understand. Please be more precise. – Bill Dubuque Nov 18 '20 at 18:39
  • I understand what you wrote about fibonacci series being a strong divisibility sequence. But Im struggling to grasp how to implement this into showing a concrete proof. I cant find any sub series in the Fibonacci numbers that all elements are co prime. I have a few guesses but they all lead me to dead end e.g. $F_{5n+1} , F_{5n-1}$ or not understanding how to prove it. If u could please show me explicitly one example I would be grateful. Thank you very much for your time. – ATB Nov 18 '20 at 18:58
  • @atb As a concrete example let's choose our induces to be $,n_k := p_k = k$'th prime, which are clearly pairwise coprime. Then by above these prime indexed fib's $,F_2,F_3,F_5,F_7,\ldots $ are also pairwise coprime. – Bill Dubuque Nov 18 '20 at 19:00
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    Thanks I understand that much better. although not all prime indices are prime numbers, for example $F_{19}=4181=37\cdot113$. so how do we know for a fact that the kth prime index will be co prime with any other prime index? In other words how do I know that this stands for all prime indices? – ATB Nov 18 '20 at 19:17
  • @ATB As emphasized above, the indices need only be pairwise $\rm\color{#c00}{co}prime$. viz. $,\color{#c00}{\gcd(p,q)=\bf 1}\Rightarrow\gcd(F_p,F_q) = F_{\color{#c00}{\gcd(p,q)}} = F_{\large\bf{\color{#c00}1}} = 1.\ $ Clearly unequal primes are coprime. $\ \ $ – Bill Dubuque Nov 19 '20 at 01:30