Three. (25 points) Given that $t$ is a root of the quadratic equation
$$
x^{2}+x-1=0
$$

If positive integers $a$, $b$, and $m$ satisfy the equation
$$
(a t+m)(b t+m)=31 m
$$

find the value of $a b$.
Since $t$ is a root of the quadratic equation
$$
x^{2}+x-1=0
$$

$t$ is an irrational number, and $t^{2}=1-t$.
From the problem, we have
$$
\begin{array}{l}
a b t^{2}+m(a+b) t+m^{2}=31 m \\
\Rightarrow a b(1-t)+m(a+b) t+m^{2}=31 m \\
\Rightarrow[m(a+b)-a b] t+\left(a b+m^{2}-31 m\right)=0 .
\end{array}
$$

Since $a, b, m$ are positive integers and $t$ is an irrational number, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
m(a+b)-a b=0, \\
a b+m^{2}-31 m=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a+b=31-m \\
a b=31 m-m^{2}
\end{array}\right.
\end{array}
$$

Therefore, $a, b$ are the two integer roots of the quadratic equation in $x$
$$
x^{2}+(m-31) x+31 m-m^{2}=0
$$

The discriminant of equation (1) is
$$
\begin{array}{l}
\Delta=(m-31)^{2}-4\left(31 m-m^{2}\right) \\
=(31-m)(31-5 m) \geqslant 0 .
\end{array}
$$

Since $a, b$ are positive integers, we have
$$
a+b=31-m>0 \text {. }
$$

Thus, $0<m \leqslant \frac{31}{5}$.
Since the discriminant $\Delta$ is a perfect square, upon verification, only $m=6$ meets the requirement.
Substituting $m=6$ yields
$$
a b=31 m-m^{2}=150 .
$$