F(n+1)F(n-1)-F(n)^2=(-1)^n
From specialfunctionswiki
Theorem
The following formula holds: $$F(n+1)F(n-1)-F(n)^2=(-1)^n,$$ where $F(n)$ denotes a Fibonacci number.
Proof
References
- {{ #if: |{{{2}}}|S.L. Basin}}{{#if: V.E. Hoggatt, Jr.|{{#if: |, {{ #if: |{{{2}}}|V.E. Hoggatt, Jr.}}{{#if: |, [[Mathematician:{{{author3}}}|{{ #if: |{{{2}}}|{{{author3}}}}}]]{{#if: |, [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]{{#if: |, [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]] and [[Mathematician:{{{author6}}}|{{ #if: |{{{2}}}|{{{author6}}}}}]]| and [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]]}}| and [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]}}| and [[Mathematician:{{{author3}}}|{{ #if: |{{{2}}}|{{{author3}}}}}]]}}| and {{ #if: |{{{2}}}|V.E. Hoggatt, Jr.}}}}|}}: [[Paper:S.L. Basin/A Primer on the Fibonacci Sequence Part I{{#if: |/Volume {{{volume}}}|}}{{#if: |/{{{edpage}}}}}|A Primer on the Fibonacci Sequence Part I{{#if: |: Volume {{{volume}}}|}}{{#if: |: {{{eddisplay}}} (1963)| ({{#if: |{{{ed}}} ed., }}1963)}}]]{{#if: |, {{{publisher}}}|}}{{#if: |, ISBN {{{isbn}}}|}}{{#if: Sum of Lucas numbers | ... (previous)|}}{{#if: L(n+1)L(n-1)-L(n)^2=5(-1)^(n+1) | ... (next)|}}{{#if: |: Entry: {{#if: |[[{{{entryref}}}|{{{entry}}}]]|{{{entry}}}}}|}}
