Show that, if $t_{3n}=x$, $t_{3n+1} = y$ and $t_{3n+2} = z$ for all values of $n$, then $p^3+q^3+3pq-1=0$

Question

A sequence of numbers $t_0,t_1,t_2,...$ satisfies $$t_{n+2} = pt_{n+1} + qt_n, ~~~n\geq0$$ where $p$ and $q$ are real. Throughout this question, $x,y,z$ are non-zero real numbers.

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 either $p+q+1=0$ or $(p-q)^2+(p+1)^2+(q+1)^2=0$. Hence show that either $x=y=z$ or $x+y+z=0$

I have attempted this question, but despite arriving at (*), I don't believe I did it in the correct way.

Here are my workings:

When $n+2 = 3p, 3p+1, 3p+2$ we get three equations.

$$\ z=py+qx\\ x=pz+qy \\ y=px+qz$$

Adding all three equations, we get, $$x+y+z = p(x+y+z) + q(x+y+z) \\ \therefore x+y+z \not=0 \implies p+q=1$$

I then cubed both sides of this equation resulting in $p^3+q^3+3p^2+3pq^2=1 \implies p^3+q^3+3pq(p+q)-1=0 \implies p^3+q^3+3pq-1=0$

then we can proceed to factor this expression to get $p+q=1$ or the sum of the squared terms.

I suspect this approach is incorrect, but not entirely sure why it's incorrect. I would appreciate an explanation on the matter.

Answer 1

The approach looks correct to me, since all the steps either follow from previous step or from given information in the question. Here I provide another solution which might be of interest for you based on linear algebra.

(FYI I am currently revising for my linear algebra exam in a few weeks, so it is a cool revision exercise for me too ;)

Firstly, we can write the linear recurrence in matrix form:

$$\begin{pmatrix} t_{n+2} \\ t_{n+1} \end{pmatrix} = \begin{pmatrix} p&q\\1&0 \end{pmatrix}\begin{pmatrix}t_{n+1}\\t_n\end{pmatrix}$$

This is usually enough. However, in this question we want to analyse the terms of period $3$, so let's extend the vectors:

$$\underbrace{\begin{pmatrix}t_{n+3}\\ t_{n+2}\\ t_{n+1}\end{pmatrix}}_{\vec{v_{n+1}}}=\underbrace{\begin{pmatrix}p&q&0\\1&0&0\\0&1&0\end{pmatrix}}_A\underbrace{\begin{pmatrix}t_{n+2}\\t_{n+1}\\t_n\end{pmatrix}}_{\vec{v_n}}$$

(Notice how the third column is all $0$, showing how useless it is.)

Now, let's look at the recurrence every $3$ terms:

$$\vec{v_{n+3}}=A^3\vec{v_n}=\begin{pmatrix} p^3+2pq& p^2q+q^2& 0\\p^2+q&pq&0\\p&q&0\end{pmatrix}\vec{v_n}$$

Since we're given $(t_{3n},t_{3n+1},t_{3n+2})=(x,y,z)$ is constant, it means that the vector $\vec{v}=\begin{pmatrix}z\\y\\x\end{pmatrix}$ satisfies

$$\vec{v} = A\vec{v}$$

What does this remind you of? Eigenvalues! This equation precisely tells us that $1$ is a eigenvalue of $A$ (and $\vec{v}$ is the corresponding eigenvector), since $\vec{v}\neq\vec{0}$.

This means that

$$\begin{align*} \det(A - I_3) &= \begin{vmatrix}p^3+2pq-1&p^2q+q^2&0\\p^2+q&pq-1&0\\p&q&-1\end{vmatrix}\\&=-\begin{vmatrix}p^3+2pq-1&p^2q+q^2\\p^2+q&pq-1\end{vmatrix} \\&= \cdots \\&= p^3+q^3-3pq-1 \\&= 0&(\text{eigenvalue})\end{align*}$$

As desired, and the rest follows.

Hope this helps!

Answer 2

For a shortcut, note that the characteristic polynomial of the linear recurrence is $\,x^2-px-q\,$, but the additional conditions also require $\,t_{n+3}=t_n\,$ with characteristic polynomial $\,x^3-1\,$. It follows that they must have a common root, so $\,0 = \text{res}(x^3-1, x^2 - p x - q)=-p^3 - 3 p q - q^3 + 1\,$.

Now back to OP's actual question.

I suspect this approach is incorrect, but not entirely sure why it's incorrect.

It is incorrect because of the assumption $x+y+z \ne 0$ made earlier, which is not given in the problem. This does not render the attempt useless, but leaves it incomplete. Going back to what's being asked:

$$ \begin{align} p^3+q^3+3pq-1=0 \\ \iff \quad\quad\quad\quad\; (p + q - 1) (p^2 - p q + p + q^2 + q + 1) = 0 \\ \iff \quad\quad (p + q - 1)\big((p-q)^2+(p+1)^2+(q+1)^2\big) = 0\tag{*} \end{align} $$

You proved that if $\,x+y+z \ne 0\,$ then $\,p+q-1=0\,$, which is sufficient for $\,(*)\,$ to hold. However, what remains to still be proved is that if $\,x+y+z = 0\,$ then either factor of $\,(*)\,$ is zero.

For that, substitute $\,z=-x-y\,$ $\, \iff t_2 = -t_1 - t_0\,$ in the recurrence relations for $\,n=0, 1\,$:

$$ \begin{cases} \begin{align} -t_1 - t_0 = t_2 &= pt_1 +q t_0 \\ \iff \quad\quad (p+1)t_1+(q+1)t_0 &= 0 \tag{1} \\ t_0 = t_3 &= pt_2 +q t_1 = p (pt_1 +q t_0 ) + q t_1 \\ \iff \quad (p^2+q)t_1+(pq-1)t_0 &= 0 \tag{2} \end{align} \end{cases} $$

Regarding $\,(1),(2)\,$ as a homogeneous linear system in $\,t_0,t_1\,$ which is known to have non-trivial solutions since $\,x,y,z \ne 0\,$, then it follows its determinant must be zero, proving $\,(*)\,$:

$$ 0 = \begin{vmatrix} p+1 & q+1 \\ p^2 + q & pq -1 \end{vmatrix} = -\big(p^2 - p q + p + q^2 + q + 1\big) $$