Partial fractions (Fibonacci system) I am going through MIT OCW course, and I suppose I've missed some point in the explanation.

So, we have the system's functional. The poles are clearly $\phi$  and $-1/\phi$ (where $\phi = 1.618\ldots$ is a golden ratio). So, we get
$$H = \frac 1 {(1-\phi R)(1+\frac 1 \phi R)}.$$
Usually I would proceed with partial fractions to get the sum of two fractions. Something similar is done here, but I don't really understand where the numerators provided in the explanation are coming from.
Any help would be appreciated. Thanks.
 A: $$
\frac 1 {(1-\varphi R)(1+\frac 1 \varphi R)} = \frac A {1-\varphi R} + \frac B {1+\frac 1 \varphi R}
$$
Multiplying both sides by the denominator that appears on the left side, we get
$$
1 = A\left( 1 + \frac 1 \varphi R \right) + B\left( 1 - \varphi R \right). \tag 1
$$
This should hold if $R$ is any number at all. So in particular, it should hold if $R=-\varphi,$ which makes the first term on the right vanish:
$$
1 = A\cdot 0 + B(1 - \varphi (-\varphi)),
$$
so
$$
1 = B(1 + \varphi^2)
$$
and so
\begin{align}
B & = \frac 1 {1 + \varphi^2} \\[10pt]
& = \frac 1 {1 + (\varphi + 1)} \text{ (Here we used the identity } \varphi^2 = \varphi + 1.) \\[10pt]
& = \frac 1 {\varphi\sqrt 5} \text{ (by the identity } 2 + \varphi = \varphi\sqrt 5.
\end{align}
Just as $R = -\varphi$ makes the first term in $(1)$ vanish, so setting $R=1/\varphi$ makes the second term vanish, and then you can proceed similarly.
So you need to use the particular nature of the number $\varphi$ to do some of the arithmetic: $$\varphi^2 = \varphi + 1 \text{ and } \varphi\sqrt 5 = \varphi + 2.$$
