Now, the Fibonacci sequence can be written as $$F_n=F_{n-1}+F_{n-2}$$ and we want to prove $$\forall n\in \mathbb{N},\ gcd(F_n,F_{n+1})=1$$
We will let this statement be $P(n)$.

(Note that the Fibonacci sequence can also be written as $F_{n+2}=F_{n+1}+F_{n}$ and we will use this to our advantage.)

For the **base case, $n=0$**:
\begin{align*}
gcd(F_n,F_{n+1})&=gcd(F_0,F_1)\\
&=gcd(0,1) \\
&=1 && \text{(Using Claim 1)}
\end{align*}
So **$P(0)$** holds.

Now, assuming **$P(k)$** holds for $n=k$, we let $g=gcd(F_k,F_{k+1})$.

Then $g \mid F_k \land g \mid F_{k+1}$ by the definition of the $gcd$(greatest common divisor).

By **Claim 2**,\begin{align*}
g \mid \overbrace{F_k + F_{k+1}}^\text{their sum} \land \ g \mid \overbrace{F_{k+1}-F_k(=F_{k-1})}^\text{their difference} && \text{(by the definition of the Fibonacci sequence)}\\
\end{align*}

So we know $ gcd(F_{k},F_{k+1})=1$ and $ gcd(F_{k-1},F_{k})=1$ by this and assuming that $P(k)$ holds.

In order to prove $P(k+1)$ for $n=k+1$, observe that:
\begin{align*}
gcd(F_n,F_{n+1})&=gcd(F_{k+1},F_{k+2})\\
&=gcd(F_{k+1},F_{k+1}+F_k)\\
&=1 && \text{(as we showed earlier using Claim 2 that the sum is also divisible)}\\
\text{Then $g=1$.}\\
\text{This proves $P(k+1)$.}
\end{align*}