8. (Maki Garden) Let $p$ be an odd prime. Determine the positive integers $x$, $y$, $x \leqslant y$, such that $\sqrt{2 p}-\sqrt{x}-\sqrt{y}$ is as small a non-negative number as possible.
Let $p=2n+1$, where $n$ is a positive integer.
First, we prove: for any positive integers $x, y, D \equiv \sqrt{2p} - \sqrt{x} - \sqrt{y}$ is never zero. Otherwise, we would have $2p = x + y + 2\sqrt{xy}$. Let $b^2, c^2$ be the largest square numbers that divide $x, y$ respectively, then $b + c \geq 2$. If we take $x = ab^2$, then we should have $y = ac^2$, and thus $2p = a(b + c)^2$. This leads to a contradiction because $2p$ cannot be divisible by any square number greater than 1. Therefore,
$$
\begin{array}{l}
D = \frac{2p - (\sqrt{x} + \sqrt{y})^2}{\sqrt{2p} + \sqrt{x} + \sqrt{y}} \\
= \frac{(2p - x - y)^2 - 4xy}{(\sqrt{2p} + \sqrt{x} + \sqrt{y})(2p - x - y + 2\sqrt{xy})}
\end{array}
$$

The numerator is a positive integer. If the numerator is greater than 1, then
$$
\frac{D \geq}{(\sqrt{4n+2} + \sqrt{x} + \sqrt{y})(4n+2 - (\sqrt{x} - \sqrt{y})^2)}.
$$

Thus,
$$
D > \frac{2}{2\sqrt{4n+2}(4n+2)} = \frac{1}{(4n+2)^{\frac{3}{2}}} > \frac{1}{16n^{\frac{3}{2}}}.
$$

The last step uses $4n+2 \leq 6n \leq 16^{\frac{2}{3}}n$. If the numerator equals 1, then $(2p - x - y)^2 = 4xy + 1$. Therefore, $2p - x - y = 2m + 1$, where $m$ is a positive integer. Let $d$ be the greatest common divisor of $m$ and $x$, and set $m = dh, x = dk$, where $h$ and $k$ are coprime. Let $g$ be the greatest common divisor of $m + 1$ and $y$, then $m + 1 = gk, y = gh$ (from $2p - x - y = 2m + 1$, we get $(2m + 1)^2 = 4xy + 1, m(m + 1) = xy$, and $h(m + 1) = ky$). Thus, $2p = x + y + m + m + 1 = (d + g)(h + k)$. Since $p$ is a prime, we must have $d = g = 1$ or $h = k = 1$. For the former, $x = k = m + 1$, and $y = h = m$, which contradicts $x \leq y$. For the latter, $x = d = m, y = g = m + 1$. Thus, $2p = x + y + 2m + 1$, which implies $p = 2m + 1$, and hence $m = n$. Therefore, $D =$
$$
\frac{1}{(\sqrt{4n+2} + \sqrt{n} + \sqrt{n+1})(2n + 1 + 2\sqrt{n(n+1)})}
$$

Thus, $D < \frac{1}{(\sqrt{4n} + \sqrt{n} + \sqrt{n})(2n + 2\sqrt{n \cdot n})}$
$$
= \frac{1}{16n^{\frac{3}{2}}}.
$$

Therefore, the smallest positive value of $D$ is obtained only when $(x, y) = \left(\frac{p-1}{2}, \frac{p+1}{2}\right)$.