10. (9th Central American and Caribbean Mathematical Olympiad) Let $S$ be a finite set of integers. Suppose that for any two distinct elements $p, q \in S$, there exist three elements $a, b, c \in S$ (not necessarily distinct, and $a \neq 0$) such that the polynomial $F(x) = a x^{2} + b x + c$ satisfies $F(p) = F(q) = 0$. Determine the maximum number of elements in $S$.

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10. It is easy to verify that $S=\{-1,0,1\}$ satisfies the condition.

Below is the proof: $|S|_{\max }=3$.
(1) At least one of $1, -1$ belongs to $S$.

Otherwise, there exist $a_{1}, a_{2} \in S$, and
$$
\left|a_{1}\right| \cdot\left|a_{2}\right| \geqslant 2\left(\left|a_{2}\right| \geqslant\left|a_{1}\right|\right) \text {. }
$$

By the problem statement, there exist $a, b, c \in S$, for $a_{1}, a_{2}$ satisfying the given condition.
$$
\text { Hence } \frac{c}{a}=a_{1} a_{2} \Rightarrow c=a a_{1} a_{2} \text {. }
$$

Then there exists $a_{3}=c \in S$, and
$$
\left|a_{3}\right|=|a|\left|a_{1}\right|\left|a_{2}\right|>\left|a_{2}\right| \geqslant\left|a_{1}\right| \text {. }
$$

Repeating the above process, we get $a_{i}(i=1,2, \cdots) \in S$, and $\left|a_{1}\right| \leqslant\left|a_{2}\right|\left|a_{1}\right|$. There exists $a_{2}=c \in S\left(\left|a_{2}\right|>\left|a_{1}\right|\right)$.
Thus, there exists $\left|a_{1}\right|\left|a_{1}\right|$. Similarly, this leads to a contradiction.
(ii) $a_{1} \leqslant-2$.

Consider $-\frac{b}{a}=a_{1}+1, \frac{c}{a}=a_{1}$.
By assumption, there does not exist $a \in S$, and $a \geqslant 2$.
Since $a_{1} \leqslant-2, b=-a\left(a_{1}+1\right), c=a a_{1}$, so $b$ and $c$ have opposite signs.
When $a \leqslant-2$, $c>\left|a_{1}\right| \geqslant 2$. Contradiction.
When $a=-1$, $b=a_{1}+1, c=-a_{1} \geqslant 2$. Contradiction.
When $a=1$, $b=-a_{1}-1, c=a_{1}$.
If $a_{1} \leqslant-3$, then $b \geqslant 2$. Contradiction.
From (i) and (ii), we know that $a_{1} \in\{-2,-1,0\}$.
Clearly, $S=\{-2,-1,0,1\}$ does not satisfy the condition.
In fact, for $-1, -2 \in S, x^{2}+3 x+2=0$, which is impossible.
Thus, $|S|_{\max }=3$.