3. Let $p=k^{2}+r$, where $k, r$ are integers, and satisfy $0 \leqslant r \leqslant 2 k$.
Since $\frac{\sqrt{p}-31}{75}$ is the fractional part of $\frac{1}{x}$, we have
$$
0 \leqslant \frac{\sqrt{p}-31}{75} < 1, \text{ and set }
$$
$$
\begin{array}{l}
\frac{1}{x}=\frac{1}{\sqrt{p}-k}=N+\frac{\sqrt{p}-31}{75}(N \geqslant 1). \\
\text { Then } \frac{\sqrt{p}+k}{r}=N+\frac{\sqrt{p}-31}{75}.
\end{array}
$$

By equating the irrational and rational parts, we get
$$
\begin{array}{l}
\frac{1}{r}=\frac{1}{75}, \frac{k}{r}=\frac{75 N-31}{75} \\
\Rightarrow r=75, k=75 N-31.
\end{array}
$$

If $N \geqslant 2$, then $k \geqslant 75 \times 2-31=119$.
Thus, $[\sqrt{p}]=k \geqslant 119$, which contradicts $31 \leqslant \sqrt{p}<106$.
Therefore, $N=1, k=44$.
Hence, $p=44^{2}+75=2011$, and 2011 is a prime number.
3. Given $p$ is a prime number, the fractional part of $\sqrt{p}$ is $x$, and the fractional part of $\frac{1}{x}$ is $\frac{\sqrt{p}-31}{75}$. Find all prime numbers $p$ that satisfy the condition.