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I've seen the property

$$f_{n+1} f_{n−1} = f_n^2 + (−1)^n, n ≥ 2.$$

for Fibonacci numbers at Abstract Algebra book of Thomas W. Judson. I've tried it for a few Fibonacci number, and I've really liked how the pattern goes. I've tried to prove it by induction, however I failed to do so. I would appreciate, if someone could prove and explain to me.

I would also like to state that, I'm not a math major. Therefore I would be really glad if you prove it as simple as possible.

Thanks.

Michael Hardy
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errorist
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    This is known as Cassini's Identity. You can find a few different proofs here: https://proofwiki.org/wiki/Cassini's_Identity – Brenton Sep 21 '15 at 21:21
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    Five proofs: http://www.cut-the-knot.org/arithmetic/algebra/CassinisIdentity.shtml – leonbloy Sep 21 '15 at 21:21

1 Answers1

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By Binet's formula (where $r=\phi, s=-1/\phi$ are the roots of $x^2-x-1=0$) and moving the $\sqrt5$ denominators,

$$5(f_{n+1}f_{n-1}-f_n^2)=(r^{n+1}-s^{n+1})(r^{n-1}-s^{n-1})-(r^{n}-s^{n})^2=2r^ns^n-r^{n+1}s^{n-1}-r^{n-1}s^{n+1}=r^ns^n\left(2-\frac rs-\frac sr\right)=(rs)^{n-1}(4rs-(r+s)^2).$$

As $-rs=r+s=1$, this equals $$(-1)^{n-1}(-5).$$