Convergence of sequence depending on initial value

I am considering the sequence $a_{n+1}=a_n^2-1$ and I want to examine for what values of $a_1\in\mathbb{R}$ the sequence converges. I know that if it converges it converges to $\frac{1\pm\sqrt{5}}{2}$. Any hints on how to approach this problem?

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To get an idea of what the answer would be, you could draw the graphs of $y=x$ and $y=x^2-1$ and then use cobwebbing with different values of $a_1$. – user84413 Aug 7 '14 at 19:31

If $f(x)=x^2-1$, then $|f^{\prime}(x)|=|2x|>1$ for $x=\frac{1\pm\sqrt{5}}{2}$, so both equilibrium points are locally unstable. Therefore the sequence will only converge for values of $a_1$ which are eventually mapped to
$\phi_{1}=\frac{1+\sqrt{5}}{2}\;$ or $\;\;\phi_{2}=\frac{1-\sqrt{5}}{2}$ by $f(x)$,
such as $\;\;\pm\phi_{1}, \pm\phi_{2}, \pm\sqrt{\phi_{1}}, \pm\sqrt{1+\phi_{1}}, \pm\sqrt{1+\sqrt{1+\phi_{1}}},\cdots$
This is not true. The two equilibrium points are unstable only rule out the case $a_n$ approach them closer and closer as a limit. It didn't rule out the case at some $n$, $a_n$ lands on one of these point exactly and then trap at that point forever. For an example, the sequence start with $a_1 = \frac{\sqrt{5}-1}{2}$ or $\sqrt{1 + \frac{\sqrt{5}-1}{2}}, \ldots$ converges to $\frac{1-\sqrt{5}}{2}$. – achille hui Aug 7 '14 at 22:56
This means that if $a_1$ is a value near one of the two possible limits, the sequence will not in general approach the limit, but will diverge instead. (Please see @achille hui's comment.) – user84413 Aug 8 '14 at 16:03