14.4. 45 * Find the prime triplets $(a, b, c)$ that satisfy the following conditions:
(1) $a<b<c<100$;
(2) $a+1, b+1, c+1$ form a geometric sequence.
Solve: There are 25 prime numbers $p$ less than 100, so $p+1 \in\{3,1,6,8,12,14,18,20$, $24,30,32,38,42,44,48,54,60,62,68,72,74,80,84,90,98\}$. Since $(b+$ $1)^{2}=(a+1)(c+1)$, we decompose $4^{2}, 6^{2}, 8^{2}, 12^{2}, \cdots, 90^{2}$ into the product of two numbers, hoping that the two factors are both in the above set. Only $6^{2}=3 \times 12,12^{2}=3 \times 48=6 \times 24=8 \times$ $18,18^{2}=6 \times 54,24^{2}=8 \times 72=12 \times 48=18 \times 32,12^{2}=18 \times 98,48^{2}=32 \times$ $72,84^{2}=72 \times 98$ meet the requirements. Therefore, there are 11 sets of prime triples, namely $(2,5,11),(2,11$, $47),(5,11,23),(5,17,53),(7,11,17),(7,23,71),(11,23,47),(17$, $23,31),(17,41,97),(31,47,71),(71,83,97)$. This problem can be solved by enumeration. When there are only a finite number of cases, the enumeration method is a commonly used approach.