Problem source

A sequence of numbers $t_0$, $t_1$, $t_2$, $\ldots\,$ satisfies
\[
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  
t_{n+2} = p t_{n+1}+qt_{n}  \ \ \ \ \ \ \ \ \ \ (n\ge0),
\]
where $p$ and $q$ are real. Throughout this question, $x$, $y$ and $z$ are non-zero real numbers.
\begin{questionparts}
\item Show that, if $t_n=x$ for all values of $n$, then $p+q=1$ and $x$ can be any (non-zero) real number.
\item  Show that, if $t_{2n} = x$ and $t_{2n+1}=y$ for all values of $n$, then $q\pm p=1$. Deduce that either $x=y$ or $x=-y$, unless $p$ and $q$ take certain values that you should identify.
\item 
Show that, if $t_{3n} = x$,  $t_{3n+1}=y$ and $t_{3n+2}=z$ for all values of $n$, then
\[ 
p^3+q^3 +3pq-1=0\,.
\]
 Deduce that either $p+q=1$ or $(p-q)^2 +(p+1)^2+(q+1)^2=0$. Hence show that either $x=y=z$ or $x+y+z=0$.
\end{questionparts}

Solution source

\begin{questionparts}
\item Suppose $t_n = x$ for all $n$, then we must have

\begin{align*}
&& x &= p x + q x \\
\Leftrightarrow && 1 &= p+q
\end{align*}
and this clearly works for any value of $x$.

\item Suppose $t_{2n} = x, t_{2n+1} = y$ for all $n$, then

\begin{align*}
&& x &= py + q x \\
&& y &= px + q y \\
\Rightarrow && 0 &= py + (q-1) x \\
&& 0 &= px + (q-1) y \\
\Rightarrow && p &= (q-1) \frac{x}{y} = (q-1) \frac{y}{x} \\
\Rightarrow && \frac{y}{x} = \pm 1 & \text{ or } q = 1, p = 0 \\
\Rightarrow && y = \pm x & \text{ or } (p,q) = (0,1) \\
\end{align*}

\item Suppose $t_{3n} = x$,  $t_{3n+1}=y$ and $t_{3n+2}=z$ , so

\begin{align*}
&& x &= pz + qy \\
&& y & = px + qz \\
&& z &= py + qx \\
\\
&& z &= p(px+qz) + q(pz + qy) \\
&&&= p^2x + 2pqz + q^2 y \\
&&&= p^2(pz+qy) + 2pqz + q^2(px+qz) \\
&&&= p^3 z + p^2qy + 2pqz + q^2p x + q^3 z \\
&&&= (p^3+q^3+2pq)z + pq(py+qx) \\
&&&= (p^3 + q^3 + 2pq)z + pq z \\
&&&=  (p^3 + q^3 + 3pq)z \\
\Rightarrow && 0 &=  p^3 + q^3 + 3pq- 1 \\
&&&= (p+q-1)(p^2+q^2+1+p+q-pq) \\
&&&= \tfrac12(p+q-1)((p-q)^2+(p+1)^2+(q+1)^2)
\end{align*}

Therefore $p+q = 1$ or $(p-q)^2+(p+1)^2+(q+1)^2 = 0 \Rightarrow p = q = -1$.

If $p+q = 1$, then $z = py + (1-p)x$ and $x = p(py+(1-p)x) + (1-p)y \Rightarrow (1-p+p^2)x = (1-p+p^2)y  \Rightarrow x = y \Rightarrow x=  y = z$.

If $p = q = -1$ then adding all the equations we get $x + y + z = -2(x+y+z) \Rightarrow x + y + z = 0$
\end{questionparts}

Note that what is actually going on here is that solutions must be of the form $t_n = \lambda^n$ so the only way to be constant is for $\lambda = 1$ to be a root, the only way for it to be $2$-periodic is for $\lambda = -1$ to be a root, and the only way for it to be $3$-periodic is for $\lambda = 1, \omega, \omega^2$ to be the roots (although we see this via the classic $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x + \omega y + \omega^2 z)(x+\omega^2 y +\omega z)$ which is because of the real constraint in the question.