# Thread: Euclid's algorithm and fibonacci numbers

1. ## Euclid's algorithm and fibonacci numbers

Let $F_{0}, F_{1}, F_{2}, F_{3},...$ be the Fibonacci numbers, defined by

$F_{0}=1$

$F_{1}=1$

$F_{n}=F_{n-1}+F_{n-2}$

for $n\geq 0$

a) Prove that for all $n\geq 0$ we have $gcd(F_{n+1}, F_{n})=1$

b) Prove that for all $n\geq 0$ we also have $gcd(F_{n+2}, F_{n})=1$

(Check what Euclid's algorithm would do if you started to compute $gcd(F_{n+1}, F_{n})$ or $gcd(F_{n+2}, F_{n})$.)

Any help would be greatly appreciated!

2. ## Re: Euclid's algorithm and fibonacci numbers

a) I would use the algorithm to state:

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

Thus, iterating the algorithm $n-1$ times, we will find $\gcd(F_{n+1},F_{n})=\gcd(F_2,F_1)=\gcd(1,1)=1$.

b) The first subtraction gives:

$\gcd(F_{n+2},F_{n})=\gcd(F_{n+1},F_{n})$ now using the result from part a) we then find:

$\gcd(F_{n+2},F_{n})=1$

3. ## Re: Euclid's algorithm and fibonacci numbers

hi
I neead an algorithm for drawing this: