<s>[INST] Find all prime triples $(a,b,c)$ such $a+1,b+1,c+1$ form a geometric sequence 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?
 [/INST] 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. 
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