The first part i) I can do.
For part ii) this is how far I can get:
If n is odd then $y=px+qy$
If n is even then $x=py+qx$
After some rearranging i end up with $y(1-q)=px$ & $x(1-q)=py$ and dividing one by the other I can show that $\frac{y}{x}=\frac{x}{y}$ from which it follows that $x^2=y^2$ and so $x=y$ or $x=-y$ as required. But how do I show that $q\pm p=1$?
For part iii) I'm afraid I have simply no idea, so any hints or advice will be greatly appreciated.
With kind regards.
Thank you very much.
Following shooting-squirrel's advice I get:
$(1-pq)x=(p^2+q)z$ [1]
$(1-pq)z=(p^2+q)y$ [2]
$(1-pq)x=(p^2+q)y$ [3]
Rearranging I get $x=\frac{(p^2+q)y}{1-pq}=\frac{(p^2+q)z}{1-pq}$ & $z=\frac{(p^2+q)y}{1-pq}$ $\implies$ $x=\frac{(p^2+q)y}{1-pq}=\frac{((p^2+q)^2)y}{(1-pq)^2}$ $\implies$ $p^2 + q=1-pq$ $\ne p^3 +q^3 +3pq-1=0$ which was what was required.
By the way this might be because you might have made a mistake with those 3 equations the ones i get are:
$z=py+qx$ [For $n=0$]
$x=pz+qy$ [For $n=1$]
$y=px+qz$ [For $n=2$]
leading to
$(1-pq)x=(q^2+p)z$ [1]
$(1-pq)z=(q^2+p)y$ [2]
$(1-pq)x=(q^2+p)y$ [3]
Part 2
$y=px+qy$
$x=py+qx$
$(1-q)y=px$
$(1-q)x=py$
If $q=1 => p=0$
Otherwise,
$y=\frac{p}{1-q}x=(\frac{p}{1-q})^2y$
$\frac{p}{1-q}=1$ or $\frac{p}{1-q}=-1 => x=y$ or $x=-y$
$\frac{p}{1-q}-1=0 => p+q=1$
$\frac{p}{1-q}+1=0 => p-q=-1$
Part 3
Our sequence, $t_{n}$ looks like x,y,z,x,y,z,x,y,z,...
Using the first rule in your problem statement(which applies to any 3 consecutive numbers in the sequence)
$x=py+qz$(#) (here $t_{n+2}=x$)
$y=pz+qx$(##) (here $t_{n+2}=y$)
$z=px+qy$(###) (here $t_{n+2}=z$)
All you need to do here, is to apply the same idea as in part 2. Use (##) to substitute for y in (#), then you'll get a relation between x and z. Now, you want to get a relation between z and y, and afterwards, between y and x. Now, use the the 3 newly acquired relations as we used them in part 2.