Catalan’s identity. Let \({F}_{n}\) denote the \(n\)th Fibonacci number with \({F}_{0}=0, {F}_{1}=1, {F}_{n+1}={F}_{n}+{F}_{n-1}\). Then the identity \({F}_{n}^{2}-{F}_{n+r}{F}_{n-r}={\left(-1\right)}^{n-r}{F}_{r}^{2}\) holds for all positive integers.
Lemma. Let \(f\left(n,r\right)={F}_{n}^{2}-{F}_{n+r}{F}_{n-r}\). Then \(f\left(n-1,r\right)+f\left(n,r\right)=0\).
That is, \({F}_{n+r}{F}_{n-r}+{F}_{n+r-1}{F}_{n-r-1}={F}_{n}^{2}+{F}_{n-1}^{2}\).
Proof. By induction.
When \(r=1\), we have \({F}_{n+1}{F}_{n-1}+{F}_{n}{F}_{n-2}={F}_{n}^{2}+{F}_{n-1}^{2}\).
Suppose that \({F}_{n+r}{F}_{n-r}+{F}_{n+r-1}{F}_{n-r-1}={F}_{n}^{2}+{F}_{n-1}^{2}\) is true when \(r=k\).
That is, \({F}_{n+k}{F}_{n-k}+{F}_{n+k-1}{F}_{n-k-1}={F}_{n}^{2}+{F}_{n-1}^{2}\).
When \(r=k+1\), we must show that \({F}_{n+k+1}{F}_{n-k-1}+{F}_{n+k}{F}_{n-k-2}={F}_{n}^{2}+{F}_{n-1}^{2}\).
Theorem (Catalan’s identity). \({F}_{n}^{2}-{F}_{n+r}{F}_{n-r}={\left(-1\right)}^{n-r}{F}_{r}^{2}.\)
Proof. Since \(f\left(n,r\right)=-f(n-1,r)\), \(f\left(r,r\right)={F}_{r}^{2}\), it follows that \(f\left(n,r\right)={\left(-1\right)}^{n-r}{F}_{r}^{2}\). That is, \({F}_{n}^{2}-{F}_{n+r}{F}_{n-r}={\left(-1\right)}^{n-r}{F}_{r}^{2}\).
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Heo, N.G. Visual Proof of Catalan’s Identity for the Fibonacci Numbers. Math Intelligencer (2024). https://doi.org/10.1007/s00283-024-10340-7
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DOI: https://doi.org/10.1007/s00283-024-10340-7