### Instruction: A sequence $\{a_n\}_{n \ge 0}$ of real numbers satisfies the recursion $a_{n+1} = a_n^3 - 3a_n^2+3$ for all positive integers $n$. For how many values of $a_0$ does $a_{2007}=a_0$?### Response: I notice that the recursion involves only one previous term, so I can try to find a fixed point, that is, a value of $a_0$ that does not change under the recursion.
If I set $a_0=a_1$, I get the equation $a_0 = a_0^3 - 3a_0^2+3$, which simplifies to $a_0^3 - 3a_0^2 - a_0 + 3 = 0$.
I can try to factor this cubic equation by looking for rational roots of the form $\pm p/q$, where $p$ is a factor of 3 and $q$ is a factor of 1.
I see that $a_0=1$ is a root, since $1^3 - 3 \cdot 1^2 - 1 + 3 = 0$.
I can use synthetic division to divide the cubic polynomial by $a_0-1$ and get a quadratic quotient.
I get $(a_0-1)(a_0^2 - 2a_0 - 3) = 0$, which factors further as $(a_0-1)(a_0-3)(a_0+1) = 0$.
So the three possible values of $a_0$ that are fixed points are $-1, 1, 3$.
However, I need to check if these values actually satisfy $a_{2007}=a_0$, or if they are just fixed points for some other values of $n$.
I can use the recursion formula to compute the first few terms of the sequence for each value of $a_0$.
If $a_0=-1$, then $a_1=-1^3 - 3 \cdot -1^2 + 3 = -1$, so $a_n=-1$ for all $n$.
If $a_0=1$, then $a_1=1^3 - 3 \cdot 1^2 + 3 = 1$, so $a_n=1$ for all $n$.
If $a_0=3$, then $a_1=3^3 - 3 \cdot 3^2 + 3 = 3$, so $a_n=3$ for all $n$.
Therefore, all three values of $a_0$ satisfy $a_{2007}=a_0$, and there are no other values that do.
So the answer is 3.