= Solution
Write $F_n=x_n$. For $n\geq1$, the recurrence and the previously proved <greatest common divisor> invariance give
$$
\gcd(F_{n+1},F_n)=\gcd(F_n+F_{n-1},F_n)=\gcd(F_{n-1},F_n).
$$
Repeating reaches $\gcd(F_1,F_0)=\gcd(1,0)=1$. This also gives the $n=0$ case directly. Applying the same invariance once more,
$$
\gcd(F_{n+2},F_n)=\gcd(F_{n+1}+F_n,F_n)=\gcd(F_{n+1},F_n)=1.
$$
Thus \b[both requested greatest common divisors equal one for every $n\geq0$]. This is the adjacent-index instance of <Fibonacci gcd reduction>, derived directly rather than assuming the stronger general formula.
Back to article page
