At most $n$ functions satisfy $f(x)=\bigl(x-f(a_1)\bigr)\bigl(x-f(a_2)\bigr)\dots\bigl(x-f(a_n)\bigr)$? Some background: I was trying to solve the functional equation $f\bigl(f(x)\bigr)=\sin x$. I realized that $f(\pi n)$ is a root of $f$ for all integers $n$, because $f\bigl(f(\pi n)\bigr)=\sin(\pi n)=0$. Thus, we can write $f$ as $f(x)=A\bigl(x-f(0)\bigr)\bigl(x-f(\pi)\bigr)\bigl(x-f(-\pi)\bigr)\dots$. The problem now is to find all functions $f$ that solve this. I couldn't even begin on this, so I decided to try letting the constant $A$ equal $1$, and to try some simpler versions first. E.g. For $f(x)=x-f(0)$ we find the unique solution $f(x)=x$; for $f(x)=\bigl(x-f(0)\bigr)\Bigl(x-f\left(\sqrt{2}\right)\Bigr)$ we find two solutions, namely $f(x)\in \left\{x^2-1,x^2-2\left(\sqrt{2}-1\right)x\right\}$.
I solved many more examples, and then conjectured that:

Let $a_i\in \mathbb{C}$ for $i=1,2,\dots,n$. Prove that the functional equation $f(x)=\bigl(x-f(a_1)\bigr)\bigl(x-f(a_2)\bigr)\dots\bigl(x-f(a_n)\bigr)$ has exactly $n$ distinct solutions.

Finally, I realized that if this is proved, then it would "follow" (I don't know if this is rigorous but I hope someone knows how to make it rigorous) that our original functional equation $f(x)=\bigl(x-f(0)\bigr)\bigl(x-f(\pi)\bigr)\bigl(x-f(-\pi)\bigr)\dots$ (I put $A=1$) has infinitely many solutions.
Thanks for reading.
 A: The first random examples I tried showed that your conjecture was
mostly false. The only thing that may still be possibly true is that
there is always at least one solution.
Let us put $b_i=f(a_i)(1\leq i\leq n)$. The conditions on $f$ can then be restated
as :
$$
b_i=\prod_{k=1}^{n} (a_i-b_k) \ (1\leq i \leq n) \ (1)
$$
The number of solutions can be LESS than $n$ : consider the case
$n=2,a_1=0,a_2=1$. Then (1) yields $(i)b_1=b_1b_2$ and
$(ii)b_2=(1-b_1)(1-b_2)$. By (i) we have $b_1=0$ 
or $b_2=1$, but $b_2=1$ is impossible because of (ii). So $b_1=0$
and hence $b_2=\frac{1}{2}$. In this case, there is exactly one solution.
The number of solutions can be MORE than $n$ : consider the case
$n=3,a_1=0,a_2=1,a_3=7$. The polynomial
$$
P=2304X^4 - 39168X^3 + 214208X^2 - 402384X + 223191
$$
has four real roots $\alpha_1\approx 0.97, \alpha_2\approx 2.02, 
\alpha_3\approx 6.98,\alpha_4\approx 7.01$, and is easily seen to be
irreducible over $\mathbb Q$.
Let $\alpha$ be any of the four roots of $P$ ; let
$$
\left\lbrace\begin{array}{lcl}
b_1 &=& \alpha, \\
b_2 &=& \frac{1152\alpha^3 - 12240\alpha^2 + 32224\alpha - 20964}{525} \\
b_3 &=& \frac{-1152\alpha^3 + 12240\alpha^2 - 32944\alpha + 26004}{645} \\
\end{array}\right.
$$
Then routine calculations show that $(x-b_1)(x-b_2)(x-b_3)$ is indeed a
solution. In this case, there are thus at least four solutions.
