BM01 2008/09 Question 5 Sequences Problem 

*Determine the sequences $a_0 , a_1 , a_2 ,\dots$ which satisfy all of the following conditions:


a) $a_{n+1} = 2a_n^2 − 1$ for every integer $n ≥ 0,$ 
b) $a_0$ is a rational number and
c) $a_i =a_j$ for some $i,j$ with $i \neq j$.
You can see it clearer here: http://www.bmoc.maths.org/home/bmo1-2009.pdf
I have managed to work out the modulus of $a_0$ must be equal to or below $1$. Indeed the $1,-1,1/2,-1/2$ and $0$ all work. I'm not familiar with sequences questions so I'm not really sure what I'm looking for in terms of a proof. I think I may also worked out that if we call $a_0$ as $a/b$ then $a+b$ and $2b$ must be square numbers since the rest of the rest of the terms can be put into the form $a/b$ (obviously not the same a and b), but also in the form $2a^2-b^2/b^2$ (again not the same a and b, sorry if I'm confusing people), since the denominator plus the numerator will be in the form $2a^2$, I thought that only $a_0$ can re-appear and so its $a+b$ and $2b$ must be square, but I'm not sure if I made a mistake, because it doesn't seem to work with some values. 
Thanks in advance for any contributions. 

You have $a_{n+1}=f(a_n)$ where $f(x)=2x^2-1$.
Since $f(x)>x$ whenever $x>1$, we must have $a_n\leq 1$ for every $n$.
Since $f(x)>1$ whenever $x<-1$, we must have $a_n\geq -1$ for every $n$.
So we have $a_n\in [-1,1]$ for every $n$. 
In particular, $a_0\in [-1,1]$, so there is a $\theta\in{\mathbb R}$
such that $a_0=\cos(\theta)$. By elementary trigonometry and induction,
one deduces $a_n=\cos(2^n\theta)$ for every $n$. 
The condition $a_i=a_j,i\neq j$ then implies that $(2^j-2^i)\theta$ is an integer
multiple of $2\pi$, so that  $\theta$ is a rational multiple 
of $\pi$.
So $\frac{\theta}{\pi}$ and $\cos(\theta)$ are both rational. It is well known
that in this case, $\theta$ must be an integer multiple of $\frac{\pi}{2}$
or $\frac{\pi}{3}$ (see, for example, the last paragraph in Quiaochu’s answer
here ).
The corresponding values for $a_0$ are $0,\pm \frac{1}{2},\pm 1$.
