10. Find all four-digit numbers $m$, such that $m<2006$, and there exists a positive integer $n$, such that $m-n$ is a prime number, and $m n$ is a perfect square.
10. From the given conditions, we know that $m-n=p$ (where $p$ is a prime number), so $m=n+p$.
Let $mn=n(n+p)=x^2$, where $x$ is a positive integer. Then, $4n^2+4pn=4x^2$, which means
$$
(2n+p)^2 - p^2 = (2x)^2.
$$

Thus, $(2n-2x+p)(2n+2x+p) = p^2$.
Since $p$ is a prime number, we have
$$
\left\{\begin{array}{l}
2n-2x+p=1, \\
2n+2x+p=p^2.
\end{array}\right.
$$

Adding the two equations gives $n=\left(\frac{p-1}{2}\right)^2$.
Consequently, $m=\left(\frac{p+1}{2}\right)^2$.
Combining $1000 \leqslant m < 2006$, we get
$$
64 \leqslant p+1 \leqslant 89.
$$

Thus, the prime number $p$ can only be $67, 71, 73, 79, 83$.
Therefore, the values of $m$ that satisfy the conditions are
$$
1156, 1296, 1369, 1600, 1764.
$$
(Provided by Neng Bin)