8. Proof: The proof is divided into three steps.
(i) Suppose $R_{1}$ and $R_{2}$ are two integers that can be written in the form $a x + b y$, where $a$ and $b$ are fixed positive integers, and $x$ and $y$ are integers. Then $k_{1} R_{1} + k_{2} R_{2} \left(k_{1}, k_{2}\right.$ are integers) can also be written in the form $a x + b y$. This is because if
$$R_{1} = a x_{1} + b y_{1}, \quad R_{2} = a x_{2} + b y_{2}$$

then
$$k_{1} R_{\mathrm{i}} + k_{2} R_{2} = a \left(k_{1} x_{1} + k_{2} x_{2}\right) + b \left(k_{1} y_{1} + k_{2} y_{2}\right)$$

Since $k_{1} x_{1} + k_{2} x_{2}$ and $k_{1} y_{1} + k_{2} y_{2}$ are both integers, $k_{1} R_{1} + k_{2} R_{2}$ is also a number of the form $a x + b y$.
(ii) According to Lemma 4 of Chapter 1, if $a$ and $b$ are positive integers, and $a > b$, then there must exist $q_{1} > 0$ such that
$$a = b q_{1} + r_{1}, \text{ and } 0 \leqslant r_{1} < b$$

Since $b > r_{1}$, there exists $q_{2} > 0$ such that
$$b = r_{1} q_{2} + r_{2}, \text{ and } 0 \leqslant r_{2} < r_{1}$$

Continuing this process, since the remainders $r_{1}, r_{2}, \cdots$ decrease successively, after a finite number of steps, the remainder will eventually be 0. Therefore, we have
$$\left.\begin{array}{cc} a \Rightarrow b q_{1} + r_{1}, & 0 < r_{1} < b \\ b = r_{1} q_{2} + r_{2}, & 0 < r_{2} < r_{1} \\ r_{1} = r_{2} q_{3} + r_{3}, & 0 < r_{3} < r_{2} \\ \cdots \cdots \cdots \cdots \\ r_{n-2} = r_{n-1} q_{n} + r_{n}, & 0 < r_{n} < r_{n-1} ; \\ r_{n-1} = r_{n} q_{n+1}, & \end{array}\right\}$$

The remainder in the last equation is 0. According to Lemma 8 of Chapter 1, from the first equation in (1) we have $(a, b) = (b, r_{1})$, from the second equation we have $(b, r_{1}) = (r_{1}, r_{2})$, from the third equation we have $(r_{1}, r_{2}) = (r_{2}, r_{3})$, and so on, until the last equation we have $(r_{n-1}, r_{n}) = r_{n}$. Therefore, we have:
$$(a, b) = (b, r_{1}) = (r_{1}, r_{2}) = \cdots = (r_{n-1}, r_{n}) = r_{n}$$

That is, the last remainder $r_{n}$ is the greatest common divisor of $\boldsymbol{a}$ and $\boldsymbol{b}$, which is the principle of the Euclidean algorithm. Since $\boldsymbol{a}$ and $\boldsymbol{b}$ are coprime, $r_{n} = 1$.
(iii) Rewrite (1) as:
$$\begin{aligned} r_{1} & = a - q_{1} b \\ r_{2} & = b - q_{2} r_{1} \\ r_{3} & = r_{1} - q_{3} r_{2} \\ & \ldots \cdots \cdots \\ r_{n} & = r_{n-2} - q_{n} r_{n-1} \end{aligned}$$

Here $q_{1}, \cdots, q_{n}$ are all integers. From the discussion in (i), since $\boldsymbol{a}$ and $\boldsymbol{b}$ themselves are numbers of the form $a x + b y$, from the first equation in (2), $r_{1}$ is a number of the form $a x + b y$; from the second equation, $b$ and $r_{1}$ are numbers of the form $a x + b y$, so $r_{2}$ is a number of the form $a x + b y$; from the third equation, $r_{1}$ and $r_{2}$ are numbers of the form $a x + b y$, so $r_{3}$ is a number of the form $a x + b y$. Continuing this, from the last equation, since $r_{n-2}$ and $r_{n-1}$ are numbers of the form $a x + b y$, $r_{n}$ is a number of the form $a x + b y$. From the discussion in (ii), we know that $r_{n} = 1$, so there must exist two integers $x$ and $y$ such that $a x + b y = 1$.