For $ n \ge 1$ , $F_1=F_2=1$
for $n \ge 3$ ,$F_n=F_{n-1}+F_{n-2}$
prove that $\gcd(F_{n+1},F_n)=1$
I'm trying to prove by induction we prove that when $n=1$, the equation holds, $\gcd(F_{1},F_1)=1$ next when $n=k$, suppose $\gcd(F_{k+1},F_k)=1$ is true, we prove $P(k+1)$ is true using induction hypothesis.
We're trying to prove that $\gcd(F_{k+1+1},F_{k+1})=1$.
We know that $F_{k+2}=F_{k+1}+F_{k}$, does this mean $\gcd(F_{k+1+1},F_{k+1})=F_k$?
or $F_{k+2} =1. F_{k+1} +F_k$?
or $F_{k+2} \equiv 1 \bmod F_{k+1}$?
then $$\gcd(F_{k+1+1},F_{k+1})=\gcd(F_{k+1}+F_k,F_{k+1}) =\gcd(F_{k+1},F_{k+2} \bmod F_{k+1})=\gcd(F_{k+1}+F_{k})=1$$
I'm not too sure about when $F_{k+2}=F_{k+1}+F_{k}$, is this have same meaning as $\gcd(F_{k+1+1},F_{k+1})=F_k$? or $F_{k+2} =1. F_{k+1} +F_k$.
Is there an alternative proof?