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I did a proof to show that:

If $m|F_n$ and $m|F_{n+1}$ then $m=1$

My approach was to show that if $m|F_n$ and $m|F_{n+1}$ then $m|F_{n-1}$ which means that $m$ divides every Fibonacci number which means it divides $F_1$ which means it divides 1.

Is there an easier way of doing this?

Bill Dubuque
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Mathematical Dummy
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    This seems like a pretty easy approach. What about about this proof seemed too complicated, so that it should be eliminated in another approach? – Elchanan Solomon Aug 29 '22 at 14:16
  • This argument shows that the claim is almost trivial. Which further simplification do you expect ? You could use that $\frac{F_{n+1}}{F_n}$ are the convergents of $[1,1,1,\cdots]$ and they are always in lowest terms , a very elegant approach , but to prove this all rigorously is surely not easier. – Peter Aug 29 '22 at 15:40
  • Another observation (clearly related): Euclid's algorithm directly shows the GCD is $1$. – Brian Tung Aug 29 '22 at 16:56

1 Answers1

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I think your approach is probably the easiest one. However, if you are familiar with the fact that $$F_{n+1}\cdot F_{n-1} - F_n^2 = \pm 1$$ then you can immediately conclude that $\gcd(F_n,F_{n+1}) = 1$ (recall that $\gcd(a,b)=1$ iff there exists $x,y \in \mathbb{Z}$ s.t $xa+yb=1$).

Bill Dubuque
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George Giapitzakis
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