Sum thing

$\begin{aligned}S_{n+1}=&x^{n+1}+y^{n+1}\\=&{\underbrace{(x^n+y^n)}_{S_n}\underbrace{(x+y)}_2}-\underbrace{xy}_4\underbrace{(x^{n-1}+y^{n-1})}_{S_{n-1}}\\=&2S_n-4S_{n-1}\end{aligned}$

$\implies 2S_n=S_{n+1}+4S_{n-1}$

$\Large\therefore 2\cdot 4=\boxed{8}$

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$\underline{\color{#D61F06}{\textbf{BONUS-1}}}$ $\begin{aligned}D_{n+1}=&x^{n+1}-y^{n+1}\\=&{\underbrace{(x^n-y^n)}_{D_n}\underbrace{(x+y)}_2}-\underbrace{xy}_4\underbrace{(x^{n-1}-y^{n-1})}_{D_{n-1}}\end{aligned}$

$\implies \large\boxed{D_{n+1}=2D_n-4D_{n-1}}$

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$\underline{\color{#D61F06}{\textbf{BONUS-2}}}$

$\Large{\begin{cases}S_n=x^n+y^n\\D_n=x^n-y^n\end{cases}}$

Adding and subtracting we get :-

$S_n+D_n=2x^n~~....(1)$

$S_n-D_n=2y^n~~....(2)$

Substituting $y=\dfrac{4}x$ in $x+y=2$ , we get :- $x^2-2x=-4\implies (x-1)^2=-3\implies x=1\pm\sqrt 3 i$

Neglecting negative sign since $\mathfrak I (\small x)>0$ . Hence $x=2\left(\dfrac{1}2+\dfrac{\sqrt 3i}2\right)=2e^{i\pi/3}$ . Hence $y=\dfrac{4}x=2e^{-i\pi/3}$ . Substituting in $(1)$ and $(2)$ , we get :

$\boxed{S_n+D_n=2^{n+1}e^{ni\pi/3}\\S_n-D_n=2^{n+1}e^{-ni\pi/3}}$

Time to get really involved. Let this serve as a warning that it is best to look for mathematical beauty, as opposed to bashing.

We can solve for x and y by substitution:

$x=\dfrac{4}{y} \rightarrow x+\dfrac{4}{x}=2$

$x^2-2x+4=0$

Solving the quadratic using the quadratic formula, $x=1\pm i\sqrt 3$ . Now, we can construct the same argument for y, resulting in $y=1\pm i\sqrt 3$ .

From simple observation, it is clear that if x takes one of these values, y must take the other, as if did not occur, the system of two equations would not be solved. As all equations considered in this solution are symmetric with respect to x and y, we can use them interchangeably. Therefore, wlog we let $x=1+i\sqrt 3$ and $y=1-i\sqrt 3$ .

Now it is clear that

$S_n=(1+i\sqrt 3)^n+(1-i\sqrt 3)^n$

To simplify calculation, we switch to polar form,

$e^{\dfrac{i\pi}{3}}=\dfrac{1}{2}+\dfrac{i\sqrt 3}{2}$

$2e^{\dfrac{i\pi}{3}}=1+i\sqrt 3$

Substituting

$S_n=\left(2e^{\dfrac{i\pi}{3}}\right)^n+\left(2e^{\dfrac{-i\pi}{3}}\right)^n$

$S_n=2^n e^{\dfrac{i\pi n}{3}}+2^n e^{\dfrac{-i\pi n}{3}}$

$S_n=2^n\left(\cos{\dfrac{n\pi}{3}}+i\sin{\dfrac{n\pi}{3}}+\cos{\dfrac{n\pi}{3}}-i\sin{\dfrac{n\pi}{3}}\right)$

$S_n=2^{n+1}\cos{\dfrac{n\pi}{3}}$

In the third step, the facts that cos is an even function and sin is an odd function were used.

With this form of $S_n$ , we now consider the equation

$qS_{n}=S_{n+1}+pS_{n-1}$

Substituting $S_n$

$q\left(2^{n+1}\cos{\dfrac{n\pi}{3}}\right)=2^{n+2}\cos{\dfrac{n\pi+\pi}{3}}+p\left(2^{n}\cos{\dfrac{n\pi-\pi}{3}}\right)$

Dividing by $2^n$

$2q\cos{\dfrac{n\pi}{3}}=2^{2}\cos{\dfrac{n\pi+\pi}{3}}+p\cos{\dfrac{n\pi-\pi}{3}}$

Using cosine addition formula

$2q\cos{\dfrac{n\pi}{3}}=4\left(\dfrac{\cos{\dfrac{n\pi}{3}}}{2} - \dfrac{\sqrt 3 \sin{\dfrac{n\pi}{3}}}{2}\right)+p\left(\dfrac{\cos{\dfrac{n\pi}{3}}}{2} +\dfrac{\sqrt 3 \sin{\dfrac{n\pi}{3}}}{2}\right)$

For both sides of the equation to be equal, the sines on the right hand side must cancel. Therefore, $p=4$ . We now have

$2q\cos{\dfrac{n\pi}{3}}=4\cos{\dfrac{n\pi}{3}}$

Resulting in $q=2$ .

$\boxed{pq=8}$

We know that $\begin{cases}\color{#3D99F6}{x+y=2}\\\color{#D61F06}{xy=4}\end{cases}$

$S_0 = x^0+y^0=2\\ S_1 = \color{#3D99F6}{x+y} = 2\\ S_2 = (\color{#3D99F6}{x+y})^2-2\color{#D61F06}{xy} = \color{#3D99F6}{2}S_1 - \color{#D61F06}{4}S_0\\ S_3 = (\color{#3D99F6}{x+y})(x^2+y^2)-\color{#D61F06}{xy}(x+y)=\color{#3D99F6}{2}S_2 - \color{#D61F06}{4}S_1\\ S_4 =(\color{#3D99F6}{x+y})(x^3+y^3)-\color{#D61F06}{xy}(x^2+y^2)=\color{#3D99F6}{2}S_3 - \color{#D61F06}{4}S_2\\ S_5 =(\color{#3D99F6}{x+y})(x^4+y^4)-\color{#D61F06}{xy}(x^3+y^3)=\color{#3D99F6}{2}S_4 - \color{#D61F06}{4}S_3$

Therefore, we can see that

$S_n = \color{#3D99F6}{2}S_{n-1} - \color{#D61F06}{4}S_{n-2}\\ S_{n+1} = \color{#3D99F6}{2}S_{n} - \color{#D61F06}{4}S_{n-1}\\ \color{#3D99F6}{2}S_{n} =S_{n+1} + \color{#D61F06}{4}S_{n-1}$

$\implies\color{#3D99F6}{p=2},\;\color{#D61F06}{q=4},\;\color{#3D99F6}{p}\color{#D61F06}{q}=\color{#3D99F6}{2}\times\color{#D61F06}{4}=\boxed{8}$

$\large \color{#3D99F6}{2} S_n=(x+y)(x^n+y^n)=x^{n+1}+y^{n+1}+yx^n+xy^n=S_{n+1}+xy(x^{n-1}+y^{n-1})=S_{n+1}+ \color{#D61F06}{4} S_{n-1}$

$\large \Rightarrow \color{#3D99F6}{p} \color{#D61F06}{q} = \color{#3D99F6}{2} \times \color{#D61F06}{4} = \large \color{#20A900}{\boxed{8}}$ .

The most elegant solution is by factorising. But here's an alternative approach.

$x, y$ are roots of $\begin{aligned} &&u^2-4u+2&=0\\ &\text{Multiplying by }u^{n-1}:&u^{n+1}-4u^n+2u^{n-1}&=0\\ &&4u^n&=u^{n+1}-2u^{n-1}\\ &\text{Substituting }u=x,y \text{ and adding}: &4(x^n+y^n)&=(x^{n+1}+y^{n+1})-2(x^{n-1}+y^{n-1}) \\ &&4S_n&=S_{n+1}-2S_{n-1}\\ &&pq&=2\cdot 4 =8\\ &\text{Similarly, by substracting:} &4D_n&=D_{n+1}-2D_{n-1}\\ \end{aligned}$

We don't even need to go into calculations if we know NEWTON'S SUMS!!!!!