Question

Show that \[{F_k}^2 - {F_{k+1}F
_{k-1}} = (-1)^{k-1}\] where \(F_k\) is the \(k\)th fibonacci number; and hence or otherwise prove that the \(gcd\) of any two consecutive fiboncacci numbers is \(1\)

Answer 1

fibonacci numbers
1, 1, 2, 3, 5, 8, 13, 21, 34, 55 etc.
no two consecutive fibonacci numbers are both even. thus meaning 2 cannot be a gcd.
does that help for the last bit

Answer 2

Thats means the even and odd numbers alternate in the sequence

Answer 3

for two consecutive numbers to have a gcd of 3 is something we have to disprove.

Answer 4

Yes, your observation proves \(2\) can never be a \(gcd\) of two consecutive fibonacci numbers

Answer 5

We have f1=1,f2=1,f3=2.... So obviously gcd(f1,f2)=1.
Suppose that gcd(fn,fn+1)=1, we will show that gcd(fn+1,fn+2)=1
Consider, gcd(fn+1,fn+2)=gcd(fn+1,fn+1+fn) because fn+2=fn+1+fn.
Then gcd(fn+1,fn+1+fn)=gcd(fn+1,fn)=1 ( gcd property )

Hence, gcd(fn,fn+1)=1 for all n>0.

found this

Answer 6

Clever induction !

Answer 7

what about part a

Answer 8

found that bit on mathsxchange btw

Answer 9

iam not sure what the GCD algorithm is exactly but