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.
 A: Already for three-term progressions it's somewhat surprising that
there are infinitely many solutions, because the usual probabilistic
guess for the expected number of solutions leads to a convergent sum:
a random number of size $M$ is a square with probability about
$M^{-1/2}$, so we're summing something like $1/(abc)$ over all three-term progressions
$(a,b,c)$, etc.  To be sure such a guess cannot account for non-random
patterns arising from polynomial identities, but it does suggest that
past a certain point such identities will be the only source of solutions.
Now a mindless exhaustive search over progressions
$(x,x+y,x+2y)$ with $0 < x,y < 10^4$ finds only the first six examples
$$
(1,7),\phantom+
(4,26),\phantom+
(15,97),\phantom+
(56,362),\phantom+
(209,1351),\phantom+
(780,5042)
$$
of an infinite family associated with the solutions
$(2,1)$, $(7,4)$, $(26,15)$, $(97,56)$, $(362,209)$, $(1351,780)$, etc.
of the Pell equation $x^2-3y^2=1$.  If it can be proved that these are
the only solutions then it will immediately follow that there are
no four-term arithmetic progressions with the same property.
But that seems like a very hard problem.
Here's the gp code; with a bound of $10^4$ it takes only
a few minutes.  One can surely do better with a more intelligent
search procedure (e.g. start by finding all solutions of $ab+1=r^2$
by factoring $r^2-1$).
H = 10^4
progsq(x,y,n) = sum(i=0,n-2,sum(j=i+1,n-1,issquare((x+i*y)*(x+j*y)+1)))
for(x=1,H,for(y=1,H,if(progsq(x,y,3)==3,print([x,y]))))

A: According to Magma, the projective closure of the variety associated to the problem (given by the equations
$$x(x-y) + 1 = z_1^2, \quad (x+y)(x-y) + 1 = z_2^2, \quad (x+2y)(x-y) + 1 = z_3^2,$$
$$(x+y)x + 1 = z_4^2, \quad (x+2y)x + 1 = z_5^2, \quad (x+2y)(x+y) + 1 = z_6^2 \quad)$$
is an irreducible surface in ${\mathbb P}^8$ with 34 isolated singularities. Since it is a complete intersection of six quadrics, it should be of general type (and it has trivial rational points with $x = y = 0$ and slightly less trivial ones with $y = 0$, so reduction methods will not work), which makes it very likely that this is a hard question.
Added later: You may want to look at Question 73346 for an explanation by Noam Elkies of the reasoning behind this.
A: Dujella has written many papers on Diophantine m-tuples, check out his webpage.
A: 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.
A: In Diophantine quadruple $a<b<c<d$, it holds $d\ge 4bc$ (see e.g. Lemma 14 in http://web.math.pmf.unizg.hr/~duje/pdf/bound.pdf ).
A: 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.
A: I think it is better to solve a more formal task.  We write the system.
$$\left\{\begin{aligned}&ab+T=x^2\\&ac+T=y^2\\&bc+T=z^2\end{aligned}\right.$$
We need to find solutions $a,b,c$ - that was an arithmetic progression.  This will help the solution of the equation Pell.
$$p^2-3s^2=T$$
Knowing any solution of the equation Pell $(p_0;s_0)$ you can always find the following formula.
$$p_1=2p_0+3s_0$$
$$s_1=p_0+2s_0$$
Having any decision - can immediately write down the solution of this system.
$$a=2s-p$$
$$b=2s$$
$$c=2s+p$$
$$x=s-p$$
$$y=s$$
$$z=s+p$$
Interesting here is that the $x,y,z$ looks like an arithmetic progression.
