From the formula F(n+1)F(n-1) - F(n)^2 = (-1)^n conclude that consecutive Fibonacci numbers are...?

Question

From the formula F(n+1) * F(n-1) - F(n)^2 = (-1)^n conclude that consecutive Fibonacci numbers are relatively prime.
F(n) is suposed to be F 'sub' n, or the nth Fibonacci number.

When numbers are relatively prime, it means that their gcd = +/- 1, AKA they have no common divisors.

I'm assuming this equation needs to be manipulated to the point where it shows
F(n) / F(n+1) = (-1)^n and F(n) / F(n-1) = (-1)^n
but this is my last homework problem and, after 6 hours of doing proofs, my brain is fried guys

The fact that F(n) = F(n+1) - F(n-1) is most likely the substitution necessary to manipulate the equation.
Any help is appreciated :)

Pauley Morph · Accepted Answer

F(n+1) * F(n-1) - F(n)^2 = (-1)^n

Let A = (-1)^n * F(n+1)
Let B = -(-1)^n * F(n)

Then

A F(n-1) + B F(n) =

(-1)^n * F(n+1) * F(n-1)  - (-1)^n F(n)^2 =

(-1)^n ( F(n+1) * F(n-1)  -  F(n)^2 ) =

(-1)^n * (-1)^n =

1

Hence F(n-1) and F(n) are relatively prime.

Hence consecutive Fibonacci numbers are relatively prime.

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From the formula F(n+1)F(n-1) - F(n)^2 = (-1)^n conclude that consecutive Fibonacci numbers are...?

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