If one root of the equation \[ x^4 - px^3 - 6x^2 + px +1 =0\] is \(\alpha\), prove that the others are \[-\frac{1}{\alpha},\frac{\alpha-1}{\alpha+1},\frac{1+\alpha}{1-\alpha}.\]

Let’s say \(x^4 - px^3 - 6x^2 + px +1 =f(x).\) If \(\alpha\) is a root of the equation \(f(x) = 0\), then \[\alpha^4 - p\alpha^3 - 6\alpha^2 + p\alpha +1 =0 \implies p = \dfrac{\alpha^4 - 6\alpha^2 + 1}{\alpha(\alpha^2 - 1)}.\]

If the four stated terms are roots of the equation, we should be able to factorise it accordingly. Hence the given equation has to be equivalent to

\[\begin{align*} \left(x - \alpha \right)\left(x + \frac{1}{\alpha}\right)\left(x + \frac{1-\alpha}{1+\alpha} \right)\left(x + \frac{\alpha+1}{\alpha-1} \right) = 0. \end{align*}\] Expanding the expression leads to the equation \[\begin{align*} x^4 - \dfrac{\alpha^4 - 6\alpha^2 + 1}{\alpha(\alpha^2 - 1)}x^3 -6x^2 + \dfrac{\alpha^4 - 6\alpha^2 + 1}{\alpha(\alpha^2 - 1)}x + 1 = 0, \end{align*}\]which is the same equation that we were given at the start, with \(p = \dfrac{\alpha^4 - 6\alpha^2 + 1}{\alpha(\alpha^2 - 1)}\), and hence the stated terms are indeed the four roots.

Hence or otherwise show that for all real values of \(p\) the equation has four real roots.

All we have to show is that \(f(x)=0\) has a real solution \(\alpha\) for any given \(p\) that is not equal to zero or \(\pm 1\). The other three roots are then \[-\frac{1}{\alpha},\frac{\alpha-1}{\alpha+1},\frac{1+\alpha}{1-\alpha}.\]

We have \(f(0) = 1,\) and \(f(1)=-4.\) Since \(f(x)\) is continuous between \(0\) and \(1\), it must have a real root, say \(\alpha\), between these values, and we are done.

There is a profounder way to look at this question. Define a recurrence relation by \(u_1 = x, u_n = \dfrac{u_n+1}{1-u_n}.\)

Thus \[u_2 = \dfrac{x+1}{1-x}, u_3 = \dfrac{\dfrac{x+1}{1-x}+1}{1-\dfrac{x+1}{1-x}} = -\dfrac{1}{x}, u_4 = \dfrac{-\dfrac{1}{x}+1}{1+\dfrac{1}{x}} = \dfrac{x-1}{x+1}, u_5 = \dfrac{\dfrac{x-1}{1+x}+1}{1-\dfrac{x-1}{1+x}}=x.\]

So the sequence is \(x, \dfrac{x+1}{1-x}, -\dfrac{1}{x},\dfrac{x-1}{x+1}, x, \ldots\) and the recurrence relation is periodic with period \(4\).

Recurrences like this are called *Lyness cycles*, after Robert Cranston Lyness.

Consider the equation \(x+ \dfrac{x+1}{1-x} -\dfrac{1}{x}+\dfrac{x-1}{x+1}=p.\)

If we replace \(x\) with \(\dfrac{x+1}{1-x},\) every term cycles on on, and we get \(\dfrac{x+1}{1-x} -\dfrac{1}{x}+\dfrac{x-1}{x+1}+x=p;\) in other words, the equation remains invariant.

What is this equation? If we put everything over a common denominator, we get \[\dfrac{x^4 - px^3 - 6x^2 + px + 1}{x(x + 1)(x - 1)}=0;\]

in other words, we arrive back at the equation we were given to solve at the start.

Thus if \(\alpha\) is one solution for the equation that is not \(0\) or \(\pm1,\) we have the other three immediately as \(\dfrac{\alpha+1}{1-\alpha}, -\dfrac{1}{\alpha}, \dfrac{\alpha-1}{\alpha+1}.\)