<s>[INST] Special arithmetic progressions involving perfect squares 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.
 [/INST] Starting from the equations in my previous answer, we get, by multiplying them in pairs,
$$(x-y)x(x+y)(x+2y) + (x-y)x + (x+y)(x+2y) + 1 = (z_1 z_6)^2\,,$$
$$(x-y)x(x+y)(x+2y) + (x-y)(x+y) + x(x+2y) + 1 = (z_2 z_5)^2\,,$$
$$(x-y)x(x+y)(x+2y) + (x-y)(x+2y) + x(x+y) + 1 = (z_3 z_4)^2\,.$$
Write $u = z_1 z_6$, $v = z_2 z_5$, $w = z_3 z_4$ and take differences to obtain
$$3 y^2 = u^2 - v^2 \qquad\text{and}\qquad y^2 = v^2 - w^2\,.$$
The variety $C$ in ${\mathbb P}^3$ described by these two equations is a smooth curve of genus 1 whose Jacobian elliptic curve is 24a1 in the Cremona database; this elliptic curve has rank zero and a torsion group of order 8. This implies that $C$ has exactly 8 rational points; up to signs they are given by $(u:v:w:y) = (1:1:1:0)$ and $(2:1:0:1)$. So $y = 0$ or $w = 0$. In the first case, we do not have an honest AP ($y$ is the difference). In the second case, we get the contradiction $abcd + ad + bc + 1 = 0$ ($a,b,c,d$ are supposed to be positive). So unless I have made a mistake somewhere, this proves that there are no such APs of length 4.
Addition: We can apply this to rational points on the surface. The case $y = 0$ gives a bunch of conics of the form 
$$x^2 + 1 = z_1^2, \quad z_2 = \pm z_1, \quad \dots, \quad z_6 = \pm z_1\,;$$
the case $w = 0$ leads to $ad = -1$ or $bc = -1$. The second of these gives $ad + 1 < 0$, and the first gives $ac + 1 = (a^2 + 1)/3$, which cannot be a square. This shows that all the rational points are on the conics mentioned above; in particular, (weak) Bombieri-Lang holds for this surface.
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