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I intend to prove that consecutive Fibonacci series terms are co-prime using only property of linear combinations, without recourse to $\gcd$ of the first two terms of the series being $1$, or any way to $\gcd$.

My analysis is based on simple fact that in a Fibonacci series, in general a term is reflected thrice, as shown below:

Let the Fibonacci series be having terms at each step of computation denoted by : $F_{n+1} = F_{n} + F_{n-1}$ with the terms being given in a sequence by:

$$ \begin{align} F_{n} =& \ F_{n-1}+ {\overbrace{F_{n-2}}^{\large last}} \\ F_{n-1} =& \ {\overbrace{F_{n-2}}^{\large middle}} + F_{n-3} \\ {\overbrace{F_{n-2}}^{\large first}} =& \ F_{n-3} + F_{n-4} \\ F_{n-3} =& \ F_{n-4} + F_{n-5} \\ & \ ... \\ F_2=& \ 2.F_1 \end{align} $$

Say, $F_{n-2}$ becomes first, middle and last terms in successive linear combinations. This means that an $even$ term will lead to first time being a 'first term' (as a sum of two odd numbers); then as 'middle term' (as a sum of itself, and $F_{n-3}$, an odd term), with sum on l.h.s., $ F_{n-1}$ being an odd number (as sum of even & odd) ; then finally as a 'last term' with the sum $F_{n}$ being again an odd number, as $F_{n} = 2.F_{n-2} (even) + F_{n-3} (odd)$.

Definitely, odd and even terms are co-prime, but if some more polished explanation is offered by the property of linear combinations.

jiten
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  • wrong, all the answers had used $\gcd$ in one way or the other. I have worded my question $therefore$ to not use $\gcd$. – jiten Dec 05 '17 at 12:27
  • I think this is impossible. Even your title is using gcd, if you say "coprime, which means $gcd(a,b)=1$. I think you cannot avoid this totally, but there are several answers using less results on gcd's. – Dietrich Burde Dec 05 '17 at 12:28
  • I feel alternate and more intuitive answers are better. Also, being co-prime is not $dependent$ on having $\gcd = 1$. It is the other way round. The approach I am asking, is not available. – jiten Dec 05 '17 at 12:29
  • This is not true: $a$ and $b$ are coprime if and only if $gcd(a,b)=1$. So it is completely the same. – Dietrich Burde Dec 05 '17 at 12:30
  • @all Down-voting kills the spirit of asking new type of question. I hope there was nothing wrong in asking it. – jiten Dec 05 '17 at 12:33
  • @DietrichBurde Thanks for the insight of providing equivalence, and none being the cause of the other. I hope this sort of situation (am not aware of ) would happen in the case of two axioms being equivalent, and as axioms so nothing more basic left. – jiten Dec 05 '17 at 12:54

4 Answers4

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Don't the identities $F_{2n}F_{2n-1}=F_{2n}^2-F_{2n-1}^2+1$ and $F_{2n+1}F_{2n}=F_{2n+1}^2-F_{2n}^2-1$ help here.

If consecutive Fibonacci numbers share a common factor the right hand side of these two equations should be divisible by a square greater than 1, which cannot be.

James Arathoon
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Alternatively, the identity $F_{n-1}F_{n+1}-F_n^2=\pm1$ shows that $1$ is an integer linear combination of $F_n$ and $F_{n+1}$, whence those two numbers are coprime.

Gerry Myerson
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I think there is a simpler solution. If $F_n$ and $F_{n-1}$ had a common divisor $d$ then $d$ would also divide $F_{n-2}=F_n - F_{n-1}$. Hence $d$ would also divide $F_{n-3}$, $F_{n-4}$ etc. Eventually we get back to $F_2$ and $F_1$ and we conclude that $d$ must be equal to 1.

If we cannot use the fact that gcd($F_1$,$F_2$)=1 then the stated property is not necessarily true.

gandalf61
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If you don't want to use given formulae:

by Bézout's lemma we're done if we can show that for all $n$ there exist integers $x,y$ such that \begin{align}xF_n+yF_{n-1} &=1 \\ (x+y)F_{n-1}+xF_{n-2}&=1 \\ (2x+y)F_{n-2}+(x+y)F_{n-3}&=1 \\ \vdots \\ ((n-2)x+(n-3)y)F_2+((n-3)x+(n-4)y)F_1&=1 \\ (2n-5)x+(2n-7)y&=1.\end{align} The last equation has solutions because $2n-5$ and $2n-7$ are coprime.

However, as it's already been noted, the notion of $\gcd$ is just hidden since coprimality is equivalent to it being $1$.

Vincenzo Oliva
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  • It is obvious that $xF_n+yF_{n-1} =1 \implies$ $ (x+y)F_{n-1}+xF_{n-2}=1$. I.e., $(x)( F_{n-1} + F_{n-2})+yF_{n-1}=1\implies (x+y)( F_{n-1}) + (x)F_{n-2})=1$. Later, $(x+y)( F_{n-2} + F_{n-3})+(x)F_{n-2}=1\implies$ $ (2x+y)( F_{n-2} + (x+y)F_{n-3})=1$. But, how it leads to final statement: $(2n-5)x + (2n-7)y =1$ is not clear. – jiten Feb 17 '21 at 03:00