Three, (25 points) Given that $a$ and $b$ are integers, and satisfy $a-b$ is a prime number, $ab$ is a perfect square. If $a \geqslant 2011$, find the minimum value of $a$.


Three, let $a-b=m$ (where $m$ is a prime number) and $ab=n^2$ (where $n$ is a positive integer).
$$
\begin{array}{l}
\text { From }(a+b)^{2}-4ab=(a-b)^{2} \\
\Rightarrow(2a-m)^{2}-4n^{2}=m^{2} \\
\Rightarrow(2a-m+2n)(2a-m-2n)=m^{2} \times 1 .
\end{array}
$$

Since $2a-m+2n$ and $2a-m-2n$ are both positive integers, and $2a-m+2n > 2a-m-2n$ (since $m$ is a prime number), we have:
$$
2a-m+2n=m^{2}, 2a-m-2n=1 .
$$

Solving these, we get $a=\frac{(m+1)^{2}}{4}, n=\frac{m^{2}-1}{4}$.
Thus, $b=a-m=\frac{(m-1)^{2}}{4}$.
Given $a \geqslant 2011$, i.e., $\frac{(m+1)^{2}}{4} \geqslant 2011$.
Considering $m$ is a prime number, we get $m \geqslant 89$.
In this case, $a \geqslant \frac{(89+1)^{2}}{4}=2025$.
When $a=2025$,
$$
m=89, b=1936, n=1980 \text {. }
$$

Therefore, the minimum value of $a$ is 2025.