Find all triples $(a,b,c)$ satisfying the following conditions:
(i) $a,b,c$ are prime numbers, where $a<b<c<100$.
(ii) $a+1,b+1,c+1$ form a geometric sequence.
is from mathematical Olympiad 2007 problem,this links have found 11 triplets, $$(2,5,11),(2,11,47),(5,11,23),(5,17,53),(7,23,71)$$ $$(a+1,b+1,c+1)=(8,12,18),(18,24,32),(18,42,98),(32,48,72),(72,84,98)$$
But How to prove why have this 11 triplets ?maybe have other?
You're looking for all triplets of prime numbers $(p,(p+1)k-1,(p+1)k^2-1 )$, with $k>1$ and $(p+1)k^2-1<100$ so $$1<k<\sqrt{\frac{101}{p+1}}$$
As $k$ is a rational number let $$k=\frac{a}{b}$$
$$b<a<b\sqrt{\frac{101}{p+1}}$$
And $b^2$ must divide $p+1$, so we can consider $b$ as the largest integer such that $b^2$ divides $p+1$ and find possible values of $a$.
$$\begin{array}{c|c|c} p&b&a\\\hline 2&1&2,3,4,5\\ 3&2&3,4,5,6,7,8,9,10\\ 5&1&2,3,4\\ 7&2& 3,4,5,6,7 \\ 11&2&3,4,5\\ 13&1&2\\ 17&3&4,5,6,7\\ 19,23&2&3,4\\ 29,37,41,61,73&1&\\ 31&4&5,6,7\\ 43&2&3\\ 47&4&5\\ 53&3&4\\ 59,67,83&2&\\ 71&6&7\\ 79&4&\\ 89&3&\\ 97&7&\\ \end{array} $$
So only 39 cases to try.