7 Find the prime triplets $(a, b, c)$ that satisfy 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. (Provided by Tao Pingsheng)
Solve according to the conditions,
$$
(a+1)(c+1)=(b+1)^{2},
$$

Let $a+1=n^{2} x, c+1=m^{2} y$, where $x, y$ do not contain square factors greater than 1, then it must be that $x=y$, because according to (1),
$$
(m n)^{2} x y=(b+1)^{2},
$$

then $m n \mid(b+1)$, let $b+1=m n \cdot w$, then (2) becomes,
$$
x y=w^{2} \text {. }
$$

If $w>1$, then there is a prime $p_{1} \mid w$, i.e., $p_{1}^{2} \mid w^{2}$, since $x, y$ do not contain square factors greater than 1, thus $p_{1}\left|x, p_{1}\right| y$. Let

then (3) becomes,
$$
x=p_{1} x_{1}, y=p_{1} y_{1}, w=p_{1} w_{1} \text {. }
$$
$$
x_{1} y_{1}=w_{1}^{2} ;
$$

If $w_{1}>1$, then there is a prime $p_{2} \mid w_{1}$, i.e., $p_{2}^{2} \mid w_{1}^{2}$, since $x_{1}, y_{1}$ do not contain square factors greater than 1, then $p_{2}\left|x_{1}, p_{2}\right| y_{1}$, let
$$
x_{1}=p_{2} x_{2}, y_{1}=p_{2} y_{2}, w_{1}=p_{2} w_{2},
$$

then (4) becomes, $x_{2} y_{2}=w_{2}^{2}, \cdots \cdots$, and so on, because the number of prime factors in (3) is finite, there is $r$, such that $w_{r}=1$, and from $x_{r} y_{r}=w_{r}^{2}$ we get, $x_{r}=y_{r}=1$, thus $x=p_{1} p_{2} \cdots p_{r}=y$, denoted as $x=y=k$, then we have
$$
\left\{\begin{array}{l}
a=k n^{2}-1 \\
b=k m n-1 \\
c=k m^{2}-1
\end{array}\right.
$$

where $1 \leqslant n5, m \geqslant 3$, we get
$$
c=m^{2}-1=(m-1)(m+1)
$$

is a composite number, which is a contradiction. Therefore, $k$ is either a prime or a product of several distinct primes (i.e., $k$ is greater than 1 and has no square factors greater than 1). We simply call this “$k$ has property $p$”.
(1) According to (6), $m \geqslant 2$. When $m=2$, then $n=1$, we have $\left\{\begin{array}{l}a=k-1, \\ b=2 k-1 \\ c=4 k-1,\end{array}\right.$, because $c3$, we get $c$ is a composite number:

If $k \equiv 2(\bmod 3)$, when $k$ is even, $k$ with property $p$ are 2,14, which respectively give $a=2-1=1, b=2 \cdot 14-1=27$ not a prime; when $k$ is odd, $k$ with property $p$ are $5,11,17,23$, which respectively give $a=k-1$ not a prime;

If $k \doteq 0(\bmod 3)$, $k$ with property $p$ are 3, 6, 15, 21, when $k=3$, it gives the solution $f_{1}=(a, b, c)=(2,5,11)$;

When $k=6$, it gives the solution $f_{2}=(a, b, c)=(5,11,23) ; k=15,21$ give $a=k-1$ not a prime;
$$
\text { If } m=3 \text {, then } n=2 \text { or } 1 \text {. When } m=3, n=2 \text {, }\left\{\begin{array}{l}
a=4 k-1, \\
b=6 k-1 \text {, because } \\
c=9 k-1,
\end{array}\right.
$$

the prime $c \leqslant 97$, we get $k \leqslant 10$, $k$ with property $p$ are $2,3,5,6,7,10$, when $k$ is odd $3,5,7$, it gives $c=9 k-1$ all composite;

When $k=6$, it gives $b=6 k-1=35$ composite; $k=10$ gives $a=$ $4 k-1=39$ composite; When $k=2$, it gives the solution $f_{3}=(a, b, c)=(7,11, 2,3,5,6,7,10$. When $k$ is odd $3,5,7$, it gives $b=3 k-1$ all composite; $k=2$ and 10 give $a=k-1$ not a prime; $k=6$ gives the solution $f_{4}=(a, b, c)=(5,17,53)$;
(2) When $m=4$, from $c=16 k-1 \leqslant 97$, we get $k \leqslant 6$, $k$ with property $p$ are $2,3,5,6$. When $k=6$, $c=16 \cdot 6-1=95$ composite;
$$
k=5 \text {, }\left\{\begin{array}{l}
a=5 n^{2}-1, \\
b=20 n-1,
\end{array} \text { because } n<m=4 \text {, then } n \text { can take } 1,2,3\right. \text {, respectively }
$$

give $a, b$ at least one not a prime;
$$
k=3 \text {, } c=48-1=47,\left\{\begin{array}{l}
a=3 n^{2}-1, \\
b=12 n-1,
\end{array} \text { because } n<m=4, \text { when } n=3\right.
$$

it gives $a, b$ composite;
$n=2$ gives the solution $f_{5}=(a, b, c)=(11,23,47)$ ；
$n=1$ gives the solution $f_{6}=(a, b, c)=(2,11,47) ; k=2$ gives $c=$ $16 k-1=31,\left\{\begin{array}{l}a=2 n^{2}-1, \\ b=8 n-1,\end{array} n<m=4\right.$, only when $n=3$ gives the solution $f_{7}=(a, b, c)=(17,23,31)$;
(3) When $m=5$, $c=25 k-1 \leqslant 97$, $k$ with property $p$ are 2,3, which respectively give $c=25 k-1$ composite;
(4) When $m=6$, $c=36 k-1 \leqslant 97$, $k$ with property $p$ is only 2, giving $c=2 \cdot 36-1=71$, at this time $\left\{\begin{array}{l}a=2 n^{2}-1, \\ b=12 n-1,\end{array}, n<m=6\right.$, only when $n=2$ gives the solution $f_{8}=(a, b, c)=(7,23,71)$;
When $n=4$ gives the solution $f_{9}=(a, b, c)=(31,47,71)$;
(5) When $m=7$, $c=49 k-1 \leqslant 97$, $k$ with property $p$ is only 2, giving $c=2 \cdot 49-1=97$, and $n<m=7,\left\{\begin{array}{l}a=2 n^{2}-1, \\ b=14 n-1,\end{array}\right.$, only when $n=3$ gives the solution $f_{10}=(a, b, c)=(17,41,97)$; When $n=6$ gives the solution $f_{11}=$ $(a, b, c)=(71,83,97)$;
(6) When $m \geqslant 8$, $c=64 k-1 \leqslant 97$, $k$ with property $p$ does not exist. Therefore, there are 11 solutions that satisfy the conditions, which are the above $f_{1}, f_{2}, \cdots, f_{11}$.