Three, (50 points) Do there exist distinct prime numbers $p, q, r, s$ such that their sum is 640, and $p^{2}+q s$ and $p^{2}+q r$ are both perfect squares? If they exist, find the values of $p, q, r, s$; if not, explain the reason.

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Given $p+q+r+s=640$, and $p, q, r, s$ are distinct prime numbers, we know that $p, q, r, s$ are all odd numbers.
Let 
\[
\left\{\begin{array}{l}
p^{2}+q s=m^{2}, \\
p^{2}+q r=n^{2}
\end{array}\right.
\]
Without loss of generality, assume $s1$, then from $m-p<n-p<n+p$, we get
\[
\begin{array}{l}
m+p=q=n-p. \\
\text{Thus, } s=m-p, r=n+p. \\
\text{Therefore, } p+q+r+s=p+q+2 q \\
=p+3 q=640.
\end{array}
\]

Since $s=m-p=q-2 p \geqslant 3$, it follows that $p \leqslant 90$.
By testing each prime number $p$ not greater than 90, we find that there is no solution in this case.
If $m-p=1$, then
\[
q s=m+p=2 p+1 \Rightarrow p=\frac{q s-1}{2}.
\]

Since $q<m+p<n+p$, we have
\[
\begin{array}{l}
q=n-p, r=n+p=2 p+q \\
\Rightarrow p+q+r+s=3 p+2 q+s \\
\quad=\frac{3(q s-1)}{2}+2 q+s=640 \\
\Rightarrow(3 q+2)(3 s+4)=3857=7 \times 19 \times 29 \\
\Rightarrow 3 s+4=19,3 q+2=7 \times 29 \\
\Rightarrow s=5, q=67 \Rightarrow p=167, r=401.
\end{array}
\]

In summary, $p=167, q=67, r=401, s=5$ or $p=167, q=67, r=5, s=401$.