Prove that there exists 3 numbers $a,b,c$ so that $P(a)=b, P(b)=c, P(c)=a$

The problem

Given that $P(x)=x^3-3x$. Prove that there exists $a \neq b \neq c$ such that $P(a)=b, P(b)=c, P(c)=a$

My ideas

So the conditions here makes it quite clear that I intend to use the Lagrange Interpolation: third degree polynomial versus creating three equations.

It is even more generous to think reversely that if we have 3 functions, then we can still add one other function of choice to re-create $P(x)=x^3-3x$. (Even though I think that function is $P(0)=0$

But still, I can't get the original problem done.

So any help is appreciated!

Solution 1:

The roots of P(x)-x are 0,2,-2. but all of them are simple roots of P(P(P(x)))-x because
$(P(P(P(x)))-x)'$
$=P'(P(P(x)))P'(P(x))P'(x)-1$
$=(P'(x))^3-1$
$=(P'(0))^3-1$ or $(P'(\pm2))^3-1$
$=(-3)^3-1$ or $9^3-1$
$\neq 0$
$P(P(P(x)))-x$ has 27 roots while $P(x)-x$ has 3 roots, so the 24 numbers which are roots of the first but not second can be partitioned into 3-cycles under the action of P.

Solution 2:

So I would like to introduce another solution to my problem, using trigonometry.

So I first notice that if I select $a= 2cos \alpha$, then \begin{align} P(a)=2cos3\alpha=b = 2cos \beta \\ P(b)=2cos3\beta=c=2cos\gamma \\P(c)=2cos3\gamma=a=2cos\alpha\end{align}

So $cos27\alpha =cos\alpha$, which means if I choose $\alpha+2k\pi=27\alpha$ then I get the results.

Many thanks to @cineel for your beautiful solution, I really appreciate that.

Also thanks @Taladris for the very helpful hints