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.



Special arithmetic progressions involving perfect squares
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*.


