Three, (50 points) A magician performs a magic trick: letting an audience member write any two numbers on a blackboard. Starting from the third number, each subsequent number is the sum of the two preceding numbers. When enough numbers are written (more than 10), the audience member tells the magician the last number, and the magician can calculate the next number. If the magician has strong calculation skills and is not cheating, do you know how he does it?


Three, let the number said by the audience be $x$, the magician only needs to calculate $\frac{1+\sqrt{5}}{2} x$, and the integer obtained by rounding to the nearest whole number is the next number.

In fact, by $a_{n}=a_{n-1}+a_{n-2}$, we know that $\left\{a_{n}\right\}$ is an increasing sequence of integers.
Let $a_{n+1}-\alpha a_{n}=\beta\left(a_{n}-\alpha a_{n-1}\right)$
$\Rightarrow a_{n+1}=(\alpha+\beta) a_{n}-\alpha \beta a_{n-1}$.
By comparison, we get $\alpha+\beta=1, -\alpha \beta=1$.
Solving gives $(\alpha, \beta)=\left(\frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}\right)$.
Thus, $a_{n+1}-\alpha a_{n}=\beta\left(a_{n}-\alpha a_{n-1}\right)$
$$
=\beta^{n-1}\left(a_{2}-\alpha a_{1}\right) \text {. }
$$

By symmetry, we get
$$
a_{n+1}-\beta a_{n}=\alpha^{n-1}\left(a_{2}-\beta a_{1}\right) \text {. }
$$

Subtracting the above two equations, we get
$$
\begin{array}{l}
a_{n}=\frac{\alpha^{n-1}\left(a_{2}-\beta a_{1}\right)-\beta^{n-1}\left(a_{2}-\alpha a_{1}\right)}{\alpha-\beta} \\
=\frac{\left(a_{2}-\beta a_{1}\right)}{\alpha(\alpha-\beta)} \alpha^{n}-\frac{\left(a_{2}-\alpha a_{1}\right)}{\beta(\alpha-\beta)} \beta^{n} .
\end{array}
$$

Given that $a_{1}$ and $a_{2}$ are positive integers, when $n \geqslant 11$,
$$
\begin{array}{l}
\left|\frac{\left(a_{2}-\alpha a_{1}\right)}{\beta(\alpha-\beta)} \beta^{n}\right| \leqslant\left|\frac{9 \alpha}{\beta(\alpha-\beta)} \beta^{n}\right| \\
=\frac{9}{\sqrt{5}}\left(\frac{\sqrt{5}-1}{2}\right)^{n-2}<0.06,
\end{array}
$$

That is, when $n \geqslant 11$, $\left|\frac{\left(a_{2}-\alpha a_{1}\right)}{\beta(\alpha-\beta)} \beta^{n}\right|$ is a number much less than 1.
$$
\begin{array}{l}
\text { Also, } a_{n}-\alpha a_{n-1} \\
=\frac{a_{2}-\beta a_{1}}{\alpha(\alpha-\beta)} \alpha^{n}-\frac{a_{2}-\alpha a_{1}}{\beta(\alpha-\beta)} \beta^{n}- \\
\alpha\left(\frac{a_{2}-\beta a_{1}}{\alpha(\alpha-\beta)} \alpha^{n-1}-\frac{a_{2}-\alpha a_{1}}{\beta(\alpha-\beta)} \beta^{n-1}\right) \\
=\frac{\alpha_{2}-\alpha a_{1}}{\beta(\alpha-\beta)} \beta^{n-1} \alpha(\alpha-\beta) \\
=\beta^{n-2}\left(\alpha_{2}-\alpha a_{1}\right) \text {, } \\
\end{array}
$$

Clearly, $\beta^{n-2}\left(\alpha a_{1}-a_{2}\right)$ is a number approaching 0 infinitely.
Therefore, the magician only needs to calculate $\frac{1+\sqrt{5}}{2} x$, and the integer obtained by rounding to the nearest whole number is the next number.