(50 points)(1) Prove that there exist infinitely many pairs of positive integers $x, y$ such that $x^{2}-101 y^{2}=-1$.
(2) Given $0.30102<\lg 2<0.30103$,
$$
0.32220<\lg 2.1<0.32221 \text{. }
$$

Determine the first 6 digits before the decimal point and the first 131 digits after the decimal point of the number $(10+\sqrt{101})^{101}$ (without using calculators or similar tools).
(1) Let $A_{2 n-1}=(\sqrt{101}-10)^{2 n-1}$, $B_{2 n-1}=(\sqrt{101}+10)^{2 n-1}(n=1,2, \cdots)$.
Then $A_{2 n-1} B_{2 n-1}=1$.
Also, let $B_{2 n-1}=x_{2 n-1}+y_{2 n-1} \sqrt{101}$, where
$x_{2 n-1}, y_{2 n-1} \in \mathbf{N}_{+}$. By the binomial theorem,
$A_{2 n-1}=-x_{2 n-1}+y_{2 n-1} \sqrt{101}$.
Substituting $A_{2 n-1}$ and $B_{2 n-1}$ into equation (1) gives
$$
x_{2 n-1}^{2}-101 y_{2 n-1}^{2}=-1 \text {. }
$$

Therefore, there are infinitely many pairs of positive integers $X=x_{2 n-1}, Y=$ $y_{2 n-1}(n=1,2, \cdots)$, such that
$$
X^{2}-101 Y^{2}=-1 \text {. }
$$
(2) From (1) we get $B_{2 n-1}-A_{2 n-1}=2 x_{2 n-1}$, i.e.,
$$
B_{2 n-1}=2 x_{n}+A_{2 n-1} \text {. }
$$

Since $2 x_{n} \in \mathbf{Z}$, $A_{2 n-1}$ is the fractional part of $B_{2 n-1}$, and $2 x_{2 n-1}$ is the integer part of $B_{2 n-1}$. Hence,
$$
\begin{array}{l} 
\frac{1}{21}<\sqrt{101}-10=\frac{1}{\sqrt{101}+10}<\frac{1}{10+10}=\frac{1}{20} \\
\Rightarrow 21^{-101}=\left(\frac{1}{21}\right)^{101}<A_{101}=(\sqrt{101}-10)^{101} \\
<\left(\frac{1}{20}\right)^{101}=20^{-101} \\
\Rightarrow-101(1+\lg 2.1)=-101 \lg 21 \\
<\lg A_{101}<101 \lg 20=-101(1+\lg 2) \\
\Rightarrow-133.54321=-101 \times 1.32221 \\
<\lg A_{101}<-101 \times 1.30102 \\
=-131.40302 \\
\Rightarrow 10^{-134} \times 10^{0.45679}<A_{101}<10^{-132} \times 10^{0.99698}
\end{array}
$$

Therefore, the 131 digits after the decimal point of $A_{101}$
are all 0. Thus, the 131 digits after the decimal point of $B_{101}=(\sqrt{101}+10)^{101}$ are all 0.
Next, we find the remainder when $2 x_{101}$ is divided by $10^{6}$.
By the binomial theorem,
$$
\begin{aligned}
& 2 x_{101} \\
= & 2\left(10^{101}+\mathrm{C}_{101}^{2} \times 101 \times 10^{99}+\cdots+\mathrm{C}_{101}^{96} \times 101^{48} \times 10^{5}+\right. \\
& \left.\mathrm{C}_{101}^{98} \times 101^{49} \times 10^{3}+\mathrm{C}_{101}^{100} \times 101^{50} \times 10\right) \\
\equiv & 2\left[\frac{101 \times 100 \times 99 \times 98 \times 97}{5 \times 4 \times 3 \times 2} \times 101^{48} \times 10^{5}+\right. \\
& \left.\frac{101 \times 100 \times 99}{3 \times 2}(100+1)^{49} \times 10^{3}+101^{11} \times 10\right]\left(\bmod 10^{6}\right) \\
\equiv & 101 \times 33 \times 49 \times 97 \times 101^{48} \times 10^{6}+ \\
& 33 \times(100+1)^{50} \times 10^{5}+2 \times 101^{51} \times 10\left(\bmod 10^{6}\right) \\
\equiv & 33 \times 10^{5}+2 \times(100+1)^{51} \times 10\left(\bmod 10^{6}\right) \\
\equiv & 3 \times 10^{5}+2\left(1+100 \times 51+100^{2} C_{51}^{2}\right) \times 10\left(\bmod 10^{6}\right) \\
\equiv & 3 \times 10^{5}+20+2000 \times 51+2 \times 10^{5} \times \frac{51 \times 50}{2}\left(\bmod 10^{6}\right) \\
\equiv & 3 \times 10^{5}+20+102000\left(\bmod 10^{6}\right) \\
\equiv & 300000+20+102000\left(\bmod 10^{6}\right) \\
\equiv & 402020\left(\bmod 10^{6}\right) .
\end{aligned}
$$

Therefore, the first 6 digits before the decimal point of $B_{101}=(\sqrt{101}+10)^{101}$ are 402020.

In summary, the first 6 digits before the decimal point of the number $(\sqrt{101}+10)^{101}$ are 402020, and the 131 digits after the decimal point are all 0.