L(n)^2-5F(n)^2=4(-1)^n
From specialfunctionswiki
Theorem
The following formula holds: $$L(n)^2-5F(n)^2=4(-1)^n,$$ where $L(n)$ denotes the $n$th Lucas number and $F(n)$ denotes the $n$th 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: F(n+m+1)=F(n+1)F(m+1)+F(n)F(m) | ... (previous)|}}{{#if: F(-n)=(-1)^(n+1)F(n) | ... (next)|}}{{#if: |: Entry: {{#if: |[[{{{entryref}}}|{{{entry}}}]]|{{{entry}}}}}|}}
