Amazing Mix of Triangular Numbers and Golden Ratio

An amazing fact involving the triangular numbers and Golden Ratio has been observed and proved by Tony Foster who graciously posted his discovery at the CutTheKnotMath facebook page.

Let $T_{x}=\displaystyle\frac{x(x+1)}{2}$ and $\displaystyle\varphi=\frac{\sqrt{5}+1}{2},$ the Golden Ratio, and $\phi=\varphi^{-1}.$ Then

$T_{(n-\phi )(n+\varphi )}+T_{(n+1+\varphi )(n+2-\varphi )}=(n+1)^4.$

Proof

As is well known, the Golden Ratio satisfies $\varphi^2-\varphi - 1=0$ and $\varphi - \phi - 1=0.$ Using these

(*)

$(n-\phi )(n+\varphi )=n^2+(\varphi-\phi)n-1=n^2+n-1$

and

\begin{align} (n+1+\varphi )(n+2-\varphi )&=(n+1)(n+2)+\varphi (n+2)-\varphi (n+1)-\varphi^2\\ &=(n+1)(n+2)+\varphi-\varphi^2\\ &=n^2+3n+2-1\\ &=n^2+3n+1. \end{align}

It remains only to prove that $T_{n^2+n-1}+T_{n^2+3n+1}=(n+1)^4.$ Indeed,

\displaystyle\begin{align} T_{n^2+n-1}+T_{n^2+3n+1}&=\frac{(n^2+n-1)(n^2+n)}{2}+\frac{(n^2+3n+1)(n^2+3n+2)}{2}\\ &=(n+1)\left(\frac{(n^2+n-1)n}{2}+\frac{(n^2+3n+1)(n+2)}{2}\right)\\ &=(n+1)\frac{(n^3+n^2-n)+(n^2+5n^2+7n+2)}{2}\\ &=(n+1)\frac{2n^3+6n^2+6n+2}{2}\\ &=(n+1)(n^3+3n^2+3n+1)\\ &=(n+1)^4. \end{align}

Tony has also observed a novel formula for triangular numbers in terms of the golden ratio:

$\displaystyle T_{n}=\frac{(n-\phi )(n+\varphi )+1}{2}.$

This follows from (*). 