2. Find the integer solutions of the following indeterminate equations:
(i) $7 x+15 y=0$.
(ii) $9 x-11 y=1$.
(iii) $17 x+40 y=280$.
(iv) $133 x-105 y=217$.
(v) $49 x-56 y+14 z=35$
2.
(i) Solution: Since $(7,15)=1$, only the constant term is zero, so the integer solutions of the equation are: $x=-15i, y=7t, t=0, \pm 1, \pm 2, \cdots$.
(ii) Solution: Since $11=9+2, 9=2 \times 4+1$, we get $1=9-2 \times 4=9-4 \times (11-9)=9 \times 5-11 \times 4$, so $x=5$, $y=4$ is a set of integer solutions of the equation. All integer solutions are
$$x=5+11t, \quad y=4+9t, \quad t=0, \pm 1, \pm 2, \cdots.$$
(iii) Solution: Since $(17,40)=1$, the equation has integer solutions. Now, solve $17u+40v=1$. Since
$$40=2 \times 17+6, \quad 17=2 \times 6+5, \quad 6=5+1,$$

we get
$$\begin{array}{l}
1=6-5=6-(17-2 \times 6)=-17+3 \times 6 \\
=-17+3 \times (40-2 \times 17)=17 \times (-7) \\
\quad+40 \times 3
\end{array}$$

Thus, $u=-7, v=3$ is a set of integer solutions of $17u+40v=1$. Let $x=280u, y=280v$, then
$$17x+40y=280(17u+40v)=280$$

So, $x=280 \times (-7)=-1960, y=280 \times 3=840$ is a set of integer solutions of $17x+40y=280$. All integer solutions are $x=-1960-40t, y=840+17t$, i.e.,
$$x=-40t, \quad y=7+17t, \quad t=0, \pm 1, \pm 2, \cdots$$
(iv) Solution: Since $133=7 \times 19, 105=7 \times 15, 217=7 \times 31$, the solutions of the equation are the same as those of
$$19x-15y=31$$

Now, solve
$$19u-15v=1$$

Since $19=15+4, 15=3 \times 4+3, 4=3+1$,
we get $1=4-3=4-(15-3 \times 4)=4 \times 4-15$
$$=4 \times (19-15)-15=19 \times 4-15 \times 5$$

So, $u=4, v=5$ is a set of integer solutions of $19u-15v=1$, and thus $x=31 \times 4=124, y=31 \times 5=155$ is a set of integer solutions of $19x-15y=31$. All integer solutions are
$$x=124+15t, \quad y=155+19t$$

i.e.,
$$x=4+15t, \quad y=3+19t, \quad t=0, \pm 1, \pm 2, \cdots$$
(v) Solution: Since $49=7 \times 7, 56=7 \times 8, 14=7 \times 2$, and $35=7 \times 5$, the solutions of the equation are the same as those of
$$7x-8y+2z=5$$

Let

then
$$\begin{array}{l}
7x-8v=t \\
t+2z=5
\end{array}$$

It is easy to see that $x=7t, y=6t$ is a set of integer solutions of $7x-8y=t$. So all integer solutions are
$$\left\{\begin{array}{l}
x=7t+8u, \\
y=6t+7u
\end{array} \quad u=0, \pm 1, \pm 2, \cdots\right.$$

And $t=1, z=2$ is a set of integer solutions of $t+2z=5$. All integer solutions are
$$\left\{\begin{array}{l}
t=1-2v, \\
z=2+v
\end{array} \quad v=0, \pm 1, \pm 2, \cdots\right.$$

Substituting the expression for $t$ into the expressions for $x, y$, we get all integer solutions of the original equation:
$$\left\{\begin{array}{l}
x=7-14v+8u \\
y=6-12v+7u \\
z=2+v
\end{array}\right.$$

where $u$ and $v$ take all integers.