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

[1] G. I. Bell [2007], A fresh look at peg solitaire, Math. Mag. 80(1), 16–28, MR2286485
 A: 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.
A: As $\sigma$ verifies the identity $\sigma^2=-\sigma+1$, you can perform the following reductions (the powers of $\sigma$ have been omitted for conciseness):
$$4+11+11+9+7+5+3+1+1+1+1-1-1,\\
7+15+9+7+5+3+1+1+1+1-1-1,\\
8+16+7+5+3+1+1+1+1-1-1,\\
8+15+5+3+1+1+1+1-1-1,\\
7+13+3+1+1+1+1-1-1,\\
6+10+1+1+1+1-1-1,\\
4+7+1+1+1-1-1,\\
3+5+1+1-1-1,\\
2+4+1-1-1,\\
2+3-1-1,\\
1+1-1,\\
0+0.
$$
A: 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.
