A problem by Lucas Chen

gcd(F n , F (n-1)) = gcd( F (n-1) + F (n-2) , f (n-1)) = gcd(F (n-1) + F (n-2) -f (n-1), f (n-1)) = gcd(F (n-2) , f (n-1)) = .... = gcd(F 2 , F_1) = gcd(1,1) = 1 (ans) :D

Fibonacci numbers are numbers that the sum of the previous number and the number equals the next number.

In expressions: $F = a, b, a + b, a + 2b, 2a + 3b, 3a + 5b, \cdots$ .

Not end yet.

Every greatest common factor of Fibonacci numbers are all $1$ because Fibonacci numbers are $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, \cdots$ , and their greatest common multiple is all $1$ .

Then the greatest common factor of consecutive Fibonacci numbers is all $1$ , too.

So the answer is $\boxed{1}$ .