Example 1 Determine whether the following indeterminate equations have non-zero solutions, and find a set of solutions if they exist:
(i) $5 x^{2}-14 y^{2}-41 z^{2}=0$;
(ii) $3 x^{2}-14 y^{2}-19 z^{2}=0$.
(i) Here $a=5, b=-14, c=-41$. By Theorem 1, we need to verify whether the congruence equations (2), (3), (4) have solutions. We have
$$s^{2} \equiv-(-14) \cdot(-41) \equiv 1(\bmod 5),$$

so (2) has a solution; $\square$
$$s^{2} \equiv-(-41) \cdot(5) \equiv-5 \equiv 3^{2}(\bmod 14),$$

so (3) has a solution; Finally, we have
$$s^{2} \equiv-(5) \cdot(-14) \equiv-12(\bmod 41)$$
41 is a prime, we calculate the Legendre symbol
$$\left(\frac{-12}{41}\right)=\left(\frac{-1}{41}\right)\left(\frac{2^{2}}{41}\right)\left(\frac{3}{41}\right)=\left(\frac{3}{41}\right)=\left(\frac{41}{3}\right)=\left(\frac{-1}{3}\right)=-1 \text {, }$$

so, (4) has no solution. Therefore, the indeterminate equation (i) has no non-zero solution.
(ii) Here $a=3, b=-14, c=-19$. By Theorem 1, we also need to verify whether the congruence equations (2), (3), (4) have solutions. We have
$$s^{2} \equiv-(-14) \cdot(-19) \equiv 1(\bmod 3),$$

so (2) has a solution; $\square$
$$s^{2} \equiv-(-19) \cdot(3) \equiv 1(\bmod 14),$$

so (3) has a solution; $\square$
$$s^{2} \equiv-3 \cdot(-14) \equiv 2^{2}(\bmod 19)$$

so (4) also has a solution. To find a specific solution for (ii), from the proof of the sufficiency of Theorem 1, we first need to factorize $a x^{2}+b y^{2}+c z^{2}$ into the product of two linear terms modulo $a, b, c$. Of course, we can follow the method in the proof, but in simple specific problems, we can factorize directly. We have
$$\begin{aligned}
-14 y^{2}-19 z^{2} & \equiv y^{2}-z^{2} \equiv(y-z)(y+z)(\bmod 3), \\
3 x^{2}-19 x^{2} & \equiv 5\left(9 x^{2}-57 z^{2}\right) \equiv 5\left(9 x^{2}-z^{2}\right) \\
& \equiv 5(3 x-z)(3 x+z) \\
& \equiv(x-5 z)(3 x+z)(\bmod 14), \\
3 x^{2}-14 y^{2} & \equiv-6\left(9 x^{2}-42 y^{2}\right) \equiv-6\left(9 x^{2}-4 y^{2}\right) \\
& \equiv-6(3 x-2 y)(3 x+2 y) \\
& \equiv(x-7 y)(3 x+2 y)(\bmod 19),
\end{aligned}$$

and thus we have (corresponding to equations (7), (8), and (9))
$$\left\{\begin{array}{l}
3 x^{2}-14 y^{2}-19 z^{2} \equiv(y-z)(y+z)(\bmod 3) \\
3 x^{2}-14 y^{2}-19 z^{2} \equiv(x-5 z)(3 x+z)(\bmod 14) \\
3 x^{2}-14 y^{2}-19 z^{2} \equiv(x-7 y)(3 x+2 y)(\bmod 19)
\end{array}\right.$$

We can, of course, find the corresponding factorization (10) as done in the proof of sufficiency, and then solve the congruence equations satisfying condition (12) (this is left to the reader). But in specific numerical problems, we can directly solve the following system of congruence equations (why):
$$\left\{\begin{array}{l}
y-z \equiv 0(\bmod 3), \\
x-5 z \equiv 0(\bmod 14), \\
x-7 y \equiv 0(\bmod 19), \\
|x| \leqslant[\sqrt{14 \cdot 19}]=16, \\
|y| \leqslant[\sqrt{3 \cdot 19}]=7, \\
|z| \leqslant[\sqrt{3 \cdot 14}]=6 .
\end{array}\right.$$

Starting with the variable $z$ with the smallest range, we can find that $z_{1}=-1, y_{1}=2$, $x_{1}=-5$ is a solution. At this time,
$$3 \cdot(-5)^{2}-14 \cdot(2)^{2}-19 \cdot(-1)^{2}=75-56-19=0 .$$

So, this is a solution to the indeterminate equation (ii). If, in the trial calculation, we obtain such a solution to (16): $z_{1}=5, y_{1}=5, x_{1}=-3$. At this time,
$$3 \cdot(-3)^{2}-14 \cdot 5^{2}-19 \cdot 5^{2}=-798=-3 \cdot(-14) \cdot(-19) \text {. }$$

By the discussion in the proof of sufficiency, we should take
$$\begin{array}{l}
x_{2}=-(-14) \cdot 5+(-3) \cdot(5)=55 \\
y_{2}=3 \cdot(-3)+5 \cdot 5=16 \\
z_{2}=5^{2}+3 \cdot(-14)=-17
\end{array}$$

It is easy to verify that this is a solution to the indeterminate equation (ii).
It should be noted that the system of congruence equations (16) is derived from equation (15), and its selection is clearly not unique. In fact, as long as we arbitrarily select one linear factor from the right side of each of the three congruences in equation (15), we can form the system of congruence equations (16). For example, instead of (16), we can take
$$\left\{\begin{array}{l}
y-z \equiv 0(\bmod 3) \\
3 x+z \equiv 0(\bmod 14) \\
3 x+2 y \equiv 0(\bmod 19) \\
|x| \leqslant 16,|y| \leqslant 7,|z| \leqslant 6
\end{array}\right.$$