TITLE: Special arithmetic progressions involving perfect squares
QUESTION [13 upvotes]: Prove that there are infinitely many positive integers $a$, $b$, $c$ that are consecutive terms of an arithmetic progression and also satisfy the condition that $ab+1$, $bc+1$, $ca+1$ are all perfect squares.
I believe this can be done using Pell's equation. What is interesting however is that the following result for four numbers apparently holds:
Claim. There are no positive integers $a$, $b$, $c$, $d$ that are consecutive terms of an arithmetic progression and also satisfy the condition that $ab+1$, $ac+1$, $ad+1$, $bc+1$, $bd+1$, $cd+1$ are all perfect squares.
I am curious to see if there is any (decent) solution.
Thanks.

REPLY [4 votes]: Michael Stoll has given a nice answer, but here is a 17th century argument.
  Let $a$, $b$, $c$, $d$ be an arithmetic progression
  with common difference $\Delta \ne 0$.
  Suppose that
  $$1 + a b = z^2_1, \quad 1 + a c = z^2_2, \qquad 1 + a d = z^2_3,$$
  $$1 + b c = z^2_4, \quad 1 + b d = z^2_5, \qquad 1 + c d = z^2_6.$$
 Consider the following quantities:
   $$A =  2 z^2_2 z^2_5  - z^2_1 z^2_6,$$
   $$B =  z_2 z_3 z_4 z_5 - \Delta \cdot z_1 z_6,$$
  $$C  =  z_1 z_3 z_4 z_6 - 2 \Delta \cdot z_2 z_5,$$
  $$D =  z_1 z_2 z_5 z_6 - 3 \Delta \cdot z_3 z_4.$$
  Then one may easily compute that
  $$A^2, B^2, C^2, D^2$$
  is an arithmetic progression. In particular, if the $z_i$ are rational, then we have
  constructed an arithmetic progression of four squares, contradicting a theorem of Fermat.
To be careful, we also have to check that this arithmetic progression of squares has non-trivial difference. A little algebra shows this can only happen (assuming $\Delta \ne 0$) if one of the following equations holds:
   $$ (1 + a d)(1 + b c) = 0,  \quad (1 + a d)(1 + b c) = 2 \Delta^2.$$
The first of these equations leads to no rational solutions (as in Michael's answer), the second says that a square ($z^2_3 z^2_4$) is equal to twice a (non-zero) square, which is also impossible.