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The series $f_n$ is defined by the following relation: $f_0=0,\,f_1=1$, $f_{n}=f_{n-1}+f_{n-2}$, for $n>1$. Show that $f_n$ and $f_{n+1}$ are coprime.

$$f_0=0,\,f_1=1,\,f_2=2,\,f_3=3,\,f_4=5,\,f_5=8,\,f_6=13$$

By induction:

For all $n>1$ let $P(n)$ be the assumption $\gcd(f_n,f_{n+1})=1$.

$$P(1) \ \text{is true},$$ $$P(1): \gcd(1,1)=1 $$

$$(V.L)_{n+1}=\gcd(f_{n+1}, f_{n+2})=\gcd(f_{n+1}, f_{n+1}+f_n) \\ =\gcd(f_{n+1}, f_n)=1=(H.L)_{n+1}$$

Now how do I show that if $d\mid f_{n+1}+f_n$ and $d\mid f_{n+1}$ that $\gcd(f_{n+1}, f_n)=1$?

Many thanks.

Bill Dubuque
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MisterButter
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1 Answers1

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Hint: $\gcd(f_{n+1}, f_n) = \gcd (f_{n+1} - f_n , f_n ) = \gcd(f_{n-1}, f_n)$.

Calvin Lin
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  • I can't seem to wrap my head around the divisibility logic, if |+1+ and |+1 then |+1- and | since gcd(+1,)=1 then d=1? – MisterButter Dec 11 '19 at 21:11
  • Do you know the rule that $ \gcd(a,b) = \gcd(a-b, b)$? If no, can you prove it? If yes, then that's what I applied, and then used the definition of $f_{n+1}$. If you continue the sequence, you eventually end up with $ \ldots = \gcd(f_2,f_1) = \gcd(1,1) = 1$. – Calvin Lin Dec 11 '19 at 21:12
  • I did not know that, thank you for bringing it to my attention and taking your valuable time to explain it! – MisterButter Dec 11 '19 at 21:17
  • Glad I could help. You can upvote and accept the answer. – Calvin Lin Dec 11 '19 at 21:19
  • Please strive to search for dupes before answering obvious FAQs. – Bill Dubuque Dec 11 '19 at 22:37