5. How many triples of natural numbers $(a, b, c)$, forming an arithmetic progression ( $a<b<c$ ), exist such that the numbers $a b+1, b c+1$ and $c a+1$ are perfect squares
Solution: Infinitely many.

Let $(2+\sqrt{3})^{n}=A_{n}+B_{n} \sqrt{3}\left(A_{n}, B_{n} \in \mathbb{N}\right)$. Then $(2-\sqrt{3})^{n}=A_{n}-B_{n} \sqrt{3}$ and $A_{n}^{2}-3 B_{n}^{2}=1$. Set $a=2 B_{n}-A_{n}, b=2 B_{n}, c=2 B_{n}+A_{n}$. Then $a, b, c$ form an arithmetic progression, $a b+1=\left(A_{n}-B_{n}\right)^{2}, b c+1=\left(A_{n}+B_{n}\right)^{2}, c a+1=B_{n}^{2}$.