Three. (25 points) Find all integers $n$ such that $2 n^{2}+1$ divides $n^{3}-5 n+5$.

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Three. (25 points) Find all integers $n$ such that $2 n^{2}+1$ divides $n^{3}-5 n+5$.
Three, since $2 n^{2}+1$ is an odd number, " $2 n^{2}+1$ divides $n^{3}-5 n+5$ " is equivalent to " $2 n^{2}+1$ divides $2 n^{3}-10 n+10$ ". Let
$$
2 n^{3}-10 n+10=\left(2 n^{2}+1\right)(n+A),
$$
where $A$ is an undetermined integer.
After expanding and rearranging, we get
$$
\begin{array}{l}
2 A n^{2}+11 n+A-10=0(A \neq 0) . \\
\text { Let } \Delta=11^{2}-8 A(A-10) \geqslant 0,
\end{array}
$$

which means $8 A^{2}-80 A \leqslant 121$.
So, $8 A^{2}-80 A \leqslant 120$, that is,
$$
(A-5)^{2} \leqslant 36 \text {. }
$$

Thus, $(A-5)^{2} \leqslant 36,|A-5| \leqslant 6$.
Hence, $-1 \leqslant A \leqslant 11$, and $A \neq 0$.
Noting that the axis of symmetry of $\Delta=-8 A^{2}+80 A+121$ is $A=5$, the table is as follows.
From this, we know that $A$ can only be $3, 7, 10$, and accordingly,
$$
\sqrt{\Delta}=17, 17, 11 \text {. }
$$

From (1), we get $n=\frac{-11 \pm \sqrt{\Delta}}{4 A}$, so,
$(A, \Delta, n)=(7,17,-1)$ or $(10,11,0)(A=3$ is discarded).
Therefore, the integers $n$ sought are 0 or -1.
(Wu Weizhao, School of Mathematics and Information Science, Guangzhou University,
510405)