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I'm working on a proof that $$\gcd(Fn, Fn-1) = 1$$ for all n greater than or equal to 0.

Here's what I have so far.

I used a proof by induction. I used $$\gcd(1,0) = 1 $$ for the base case.

For the inductive step I assumed

$$\gcd(F_n, F_{n-1}) = 1$$

and set up $$\gcd(F_{n+1}, F_n)$$ $$ = \gcd(F_n, F_{n+1} \bmod Fn)$$ $$ = \gcd(F_n, (F_n + F_{n-1]) \bmod Fn)$$

I'm a bit confused about what to do at this step. Do I use $$\gcd(a+b, b) = \gcd(a,b)$$? If so, how do I prove that? Would appreciate some help!

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  • $\begingroup$ "Do I use gcd(a+b,b)=gcd(a,b)" $\gcd(F_n, F_{n+1})= \gcd(F_n, F_{n+1} - F_n)=\gcd(F_n,F_{n-1})$ so I'd say you are complete done! Are you asking how to prove $\gcd(a+b, b) =\gcd(a,b)$? It's a well known result. To prove it just consider that if anything divides $b$ then it will divide $a+b$ if and only it divides $a$. $\endgroup$ – fleablood Sep 23 '20 at 23:36
  • $\begingroup$ See this question for proof of $\,\gcd(a+nb,b) = \gcd(a,b)\ \ $ $\endgroup$ – Bill Dubuque Sep 24 '20 at 0:12