Verifying a long polynomial equation in (the reciprocal of) the Golden Ratio

Question

I'm trying to show that the following equation holds true:

$$4\sigma^{12}+11\sigma^{11}+11\sigma^{10}+9\sigma^9+7\sigma^8+5\sigma^7+3\sigma^6+\sigma^5+\sigma^4+\sigma^3+\sigma^2+\sigma = 1 + 2\sigma$$

where $\sigma$ is the reciprocal of the golden ratio; that is, $\sigma := \frac12(\sqrt{5} - 1)$.

There must be a good way to show this. However, so far all my attempts have failed.

This problem has a real world application. If you are interested in background take a look at A Fresh Look at Peg Solitaire^[1] and in particular at Figure 9.

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[1] G. I. Bell [2007], A fresh look at peg solitaire, Math. Mag. 80(1), 16–28, MR2286485

Answer 1

$$\begin{align} &\quad 4 s^{12} + 11 s^{11} + 11 s^{10} + 9 s^9 + 7 s^8 + 5 s^7 + 3 s^6 + s^5 + s^4 + s^3 + s^2 - s - 1 \\ =&\quad 4 s^{12}+ 4s^{11}-4 s^{10} \\ &\quad \phantom{4 s^{10}} + 7 s^{11}+7s^{10} - 7 s^9\\ &\quad \phantom{4 s^{10} + 7 s^{11}}+8s^{10} + 8 s^9 - 8 s^8\\ &\quad\phantom{4 s^{10}+7 s^{11}+7s^{10} }+ 8s^9+8 s^8 - 8 s^7 \\ &\quad\phantom{4 s^{10}+7 s^{11}+7s^{10} + 2s^9}+7 s^8 + 7 s^7 - 7 s^6 \\ &\quad\phantom{4 s^{10}+7 s^{11}+7s^{10} + 2s^9+5 s^8} +6 s^7 +6 s^6 - 6 s^5 \\ &\quad\phantom{4 s^{10}+7 s^{11}+7s^{10} + 2s^9+5 s^8 - 2 s^7}+ 4 s^6 + 4 s^5 - 4 s^4 \\ &\quad\phantom{4 s^{10}+7 s^{11}+7s^{10} + 2s^9+5 s^8 - 2 s^7+ 6 s^6} + 3 s^5 + 3 s^4 - 3 s^3 \\ &\quad\phantom{4 s^{10}+7 s^{11}+7s^{10} + 2s^9+5 s^8 - 2 s^7 - 2 s^6 + 3 s^5} + 2 s^4 +2 s^3 - 2 s^2 \\ &\quad\phantom{4 s^{10}+7 s^{11}+7s^{10} + 2s^9+5 s^8 - 2 s^7 - 2 s^6 + 3 s^5 + 8 s^4} +2 s^3 +2 s^2 - 2 s \\ &\quad\phantom{4 s^{10}+7 s^{11}+7s^{10} + 2s^9+5 s^8 - 2 s^7 - 2 s^6 + 3 s^5 - 2 s^4 - 2 s^3} + 1 s^2 + 1 s -1 \\ =&\quad\left( s^2 + s - 1 \right)\left(\cdots\right) \\ =&\quad 0 \end{align}$$

Answer 2

Let $t={\large{\frac{\sqrt{5}-1}{2}}}$.

Then $t$ is a root of the quadratic polynomial $p(x) = x^2 + x -1$.

Let $f(x) = 4x^{12}+11x^{11}+11x^{10}+9x^9+7x^8+5x^7+3x^6+x^5+x^4+x^3+x^2+x$.

Then, using a CAS, polynomial long division of $f$ by $p$ yields $$f(x) = q(x)p(x) + r(x)$$ where $q,r$ are given by \begin{align*} q(x) &= 4x^{10}+7x^9+8x^8+8x^7+7x^6+6x^5+4x^4+3x^3+2x^2+2x+1\\[4pt] r(x) &= 2x + 1\\[4pt] \end{align*} Hence, since $p(t) = 0$, we get $$f(t) = q(t)p(t) + r(t) = q(t)(0) + r(t) = r(t) = 2t+1$$ as was to be shown.

Answer 3

As $\sigma$ verifies the identity $\sigma^2=-\sigma+1$, you can perform the following reductions (the powers of $\sigma$ have been omitted for conciseness):

$$4+11+11+9+7+5+3+1+1+1+1-1-1,\\ 7+15+9+7+5+3+1+1+1+1-1-1,\\ 8+16+7+5+3+1+1+1+1-1-1,\\ 8+15+5+3+1+1+1+1-1-1,\\ 7+13+3+1+1+1+1-1-1,\\ 6+10+1+1+1+1-1-1,\\ 4+7+1+1+1-1-1,\\ 3+5+1+1-1-1,\\ 2+4+1-1-1,\\ 2+3-1-1,\\ 1+1-1,\\ 0+0. $$

Answer 4

Work with $x$ instead of $\sigma$. You have $$x^2=-x+1$$ so $$x^3=x-x^2=+2x-1$$ $$x^4=x^2-x^3=-3x+2$$ and you will find (without surprise) that the Fibonacci numbers appear, so you can write down $$x^5=+5x-3, x^6=-8x+5 \dots$$ and this may be an efficient way of reducing to a linear expression.