Example 1 Try to find the triple $(a, b, c)$ that satisfies the following conditions:
(1) $a<b<c<100$ ($a, b, c$ are prime numbers);
(2) $a+1, b+1, c+1$ form a geometric sequence.
(4th China Southeast Mathematical Olympiad)
```
        Solution: From condition (2), we have
        (a+1)(c+1)=(b+1)^2.
```
$$
\text { Let } a+1=n^{2} x, c+1=m^{2} y \text {, where } x \text { and } y \text { do not }
$$
```
contain any square factors greater than 1.
    Then it must be that } x = y \text{. This is because from equation (1) we get
```
Thus, $m n \mid (b+1)$.
Let $b+1 = m n \cdot w$.
Then, equation (2) becomes $x y = w^{2}$.
If $w > 1$, then there exists a prime $p_{1}$ such that $p_{1} \mid w$, i.e., $p_{1}^{2} \mid w^{2}$.
Since $x$ and $y$ do not contain any square factors greater than 1, it follows that
$p_{1} \mid x$ and $p_{1} \mid y$.
Let $x = p_{1} x_{1}$, $y = p_{1} y_{1}$, and $w = p_{1} w_{1}$.
Then, from $x y = w^{2}$, we get $x_{1} y_{1} = w_{1}^{2}$.
If $w_{1} > 1$, then there exists a prime $p_{2}$ such that $p_{2} \mid w_{1}$, i.e., $p_{2}^{2} \mid w_{1}^{2}$.
Since $x_{1}$ and $y_{1}$ do not contain any square factors greater than 1, it follows that
$p_{2} \mid x_{1}$ and $p_{2} \mid y_{1}$.
Let $x_{1} = p_{2} x_{2}$, $y_{1} = p_{2} y_{2}$, and $w_{1} = p_{2} w_{2}$.
Then, from $x_{1} y_{1} = w_{1}^{2}$, we get $x_{2} y_{2} = w_{2}^{2}$.
Continuing this process, since the number of prime factors in $w$ in $x y = w^{2}$ is finite, there exists an $r$ such that $w_{r} = 1$.
From $x_{r} y_{r} = w_{r}^{2}$, we get $x_{r} = y_{r} = 1$.
Thus, $x = p_{1} p_{2} \cdots p_{r} = y$.
Let $x = y = k$. Then,
$$
\left\{\begin{array}{l}
a = k n^{2} - 1, \\
b = k m n - 1, \\
c = k m^{2} - 1.
\end{array}\right.
$$
If $m \geq 3$, then $c = m^{2} - 1 = (m-1)(m+1)$ is a composite number, which is a contradiction.
Therefore, $k$ must be a prime or a product of distinct primes (i.e., $k > 1$ and $k$ has no square factors greater than 1). We will call this property "property $P$".
From condition (3), we have:
(i) If $m = 2$, then
$$
n = 1, \left\{\begin{array}{l}
a = k - 1, \\
b = 2 k - 1, \\
c = 4 k - 1.
\end{array}\right.
$$
Since $c \leq 97$, $c$ is a composite number.
If $k \equiv 2 \pmod{3}$:
When $k$ is even, the values of $k$ with property $P$ are 2 and 14, which give
$$
a = 2 - 1 = 1, b = 2 \times 14 - 1 = 27
$$
which are not prime;
When $k$ is odd, the values of $k$ with property $P$ are 5, 11, 17, and 23, which give $a = k - 1$ that are not prime.
If $k \equiv 0 \pmod{3}$, the values of $k$ with property $P$ are 3, 6, 15, and 21:
For $k = 3$, we get the solution
$f_{1} = (a, b, c) = (2, 5, 11)$;
For $k = 6$, we get the solution
$$
f_{2} = (a, b, c) = (5, 11, 23);
$$
For $k = 15$ and 21, the values of $a = k - 1$ are not prime.
(ii) If $m = 3$, then $n = 2$ or 1.
For $m = 3$ and $n = 2$, we have
$$
\left\{\begin{array}{l}
a = 4 k - 1, \\
b = 6 k - 1, \\
c = 9 k - 1.
\end{array}\right.
$$
Since the prime $c \leq 97$, we have $k \leq 10$.
The values of $k$ with property $P$ are 2, 3, 5, 6, 7, and 10.
For odd $k = 3, 5, 7$, the values of $c = 9 k - 1$ are composite;
For $k = 6$, the value of $b = 6 k - 1 = 35$ is composite;
For $k = 10$, the value of $a = 4 k - 1 = 39$ is composite;
For $k = 2$, we get the solution
$f_{3} = (a, b, c) = (7, 11, 17)$.
The values of $k$ with property $P$ are 2, 3, 5, 6, 7, and 10.
For odd $k = 3, 5, 7$, the values of $b = 3 k - 1$ are composite;
For $k = 2, 10$, the values of $a = k - 1$ are not prime; for $k = 6$, we get the solution
$f_{4} = (a, b, c) = (5, 17, 53)$.
(iii) For $m = 4$, from $c = 16 k - 1 \leq 97$, we get $k \leq 6$.
The values of $k$ with property $P$ are 2, 3, 5, and 6.
For $k = 6$, $c = 16 \times 6 - 1 = 95$ is composite.
For $k = 5$, we have
$$
\left\{\begin{array}{l}
a = 5 n^{2} - 1, \\
b = 20 n - 1.
\end{array}\right.
$$
Since $n < m = 4$, $n$ can be 1, 2, or 3, and for each, at least one of $a$ or $b$ is not prime.
For $k = 3$, $c = 48 - 1 = 47$, and
$$
\left\{\begin{array}{l}
a = 3 n^{2} - 1, \\
b = 12 n - 1.
\end{array}\right.
$$
Since $n < m = 4$,
For $n = 3$, the values of $a$ and $b$ are composite;
For $n = 2$, we get the solution
$f_{5} = (a, b, c) = (11, 23, 47)$;
For $n = 1$, we get the solution
$f_{6} = (a, b, c) = (2, 11, 47)$.
For $k = 2$,
$c = 16 k - 1 = 31$, and
$$
\left\{\begin{array}{l}
a = 2 n^{2} - 1, \\
b = 8 n - 1.
\end{array}\right.
$$
Since $n < m = 4$, only for $n = 3$ do we get the solution
$f_{7} = (a, b, c) = (17, 23, 31)$.
(iv) For $m = 5$, $c = 25 k - 1 \leq 97$, the values of $k$ with property $P$ are 2 and 3, which give $c = 25 k - 1$ as composite.
(v) For $m = 6$, $c = 36 k - 1 \leq 97$, the value of $k$ with property $P$ is only 2, giving $c = 2 \times 36 - 1 = 71$.
At this point,
$$
\left\{\begin{array}{l}
a = 2 n^{2} - 1, \\
b = 12 n - 1.
\end{array}\right.
$$
Since $n < m = 6$, only for $n = 2$ do we get the solution
$f_{8} = (a, b, c) = (7, 23, 71)$;
For $n = 4$, we get the solution
$f_{9} = (a, b, c) = (31, 47, 71)$.
(vi) For $m = 7$, $c = 49 k - 1 \leq 97$, the value of $k$ with property $P$ is only 2, giving $c = 2 \times 49 - 1 = 97$.
Since $n < m = 7$,
$$
\left\{\begin{array}{l}
a = 2 n^{2} - 1, \\
b = 14 n - 1,
\end{array}\right.
$$
only for $n = 3$ do we get the solution
$f_{10} = (a, b, c) = (17, 41, 97)$;
For $n = 6$, we get the solution
$f_{11} = (a, b, c) = (71, 83, 97)$.
(vii) For $m \geq 8$, $c = 64 k - 1 \leq 97$, there are no values of $k$ with property $P$.
In summary, there are 11 solutions that satisfy the conditions, which are the aforementioned $f_{1}, f_{2}, \cdots, f_{11}$.
```