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.


  • $\begingroup$ //Apologies if the tags are not appropriate; didn't know where elementary nt fits better (if it fits at all on this website) $\endgroup$ – Cosmin Pohoata Feb 11 '12 at 19:05
  • $\begingroup$ What do you mean by "apparently holds"? Have you checked it computationally, or can you derive it from the three-term progression case? $\endgroup$ – Seva Feb 11 '12 at 19:12
  • $\begingroup$ I checked it computationally, sorry for not mentioning. $\endgroup$ – Cosmin Pohoata Feb 11 '12 at 19:24
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    $\begingroup$ I added the "diophantine-equations" tag. See also my comment on duje's answer below relating to the case of triples in AP. $\endgroup$ – Michael Stoll Feb 22 '12 at 19:31
  • $\begingroup$ math.stackexchange.com/questions/1607501/… $\endgroup$ – individ Jan 11 '16 at 12:03

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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    $\begingroup$ @Michael: Very nice! Let $X$ denote the surface given by your earlier answer. Then if I understand correctly, you've given a dominant rational map $X\to C$, where $C$ has genus 1 and $#C(\mathbb{Q})=8$. It seems a little surprising that a surface of general type should map onto a genus 1 curve, although I guess the fact that $X$ is an intersection of quadrics gives it a chance. But did you have some a priori reason to suspect that $X$ admitted such a covering? $\endgroup$ – Joe Silverman Feb 12 '12 at 12:52
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    $\begingroup$ @Joe: I didn't have some good a priori reason to suspect that $X$ would map to a genus 1 curve, but the observation that the analogous problem for five terms admits such a map (to the "four squares in AP" curve) provided some motivation to try a little bit longer. $\endgroup$ – Michael Stoll Feb 12 '12 at 13:13
  • $\begingroup$ Bravo! The identities look a bit less mysterious if the arithmetic progression is written symmetrically as $(x-3y,x-y,x+y,x+3y)$, when the products $(z_i z_{7-i})^2$ are invariant under $y \leftrightarrow -y$ (Then $u^2-v^2$ and $v^2-w^2$ are $12y^2$ and $4y^2$, which comes to the same curves over ${\bf Q}$.) $\endgroup$ – Noam D. Elkies Feb 12 '12 at 19:50
  • $\begingroup$ Can magma (or other software) detect the nontrivial $H^1$ of this surface which makes this analysis possible? $\endgroup$ – Noam D. Elkies Feb 13 '12 at 1:16
  • $\begingroup$ Hm. I guess you would want the $H^1$ of its desingularization, which might be a fairly complicated object. As damiano pointed out to me by email, the fact that there is a map to an elliptic curve must be related to the singularities being sufficiently bad (otherwise the surface would be simply connected). $\endgroup$ – Michael Stoll Feb 15 '12 at 11:10

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)))
  • $\begingroup$ Isn't there a similar problem involving 1,3,8,120, that says something like there is no fifth member such that the product of any two is one less than a square? It might be also that there are finitely many quadruplets, with the nontrivial ones being not arithmetic progressions. Gerhard "Ask Me About Obscure Puzzles" Paseman, 2012.02.11 $\endgroup$ – Gerhard Paseman Feb 11 '12 at 20:37
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    $\begingroup$ I was reminded of this 1,3,8,120 puzzle too, because Martin Gardner wrote about it in one of his columns decades ago and reported that the nonexistence of a fifth positive integer was proved only with difficulty (including several 1000+ digit computations, back when that was impressive). But there are infinitely many such integer quadruplets. A few Google searches turned up vixra.org/pdf/0907.0024v1.pdf with citations going as far as Euler and Diophantus! But naturally none of these are four-term arithmetic progressions... $\endgroup$ – Noam D. Elkies Feb 11 '12 at 21:43
  • $\begingroup$ Thanks for the reply! This is essentially what I did to claim that I "checked it manually": I verified the correspondence between the solutions of the problem in question and $x^{2}-3y^{2}=1$ up to $10^{9}$ and then just assummed this to be true. I was hoping however to find some slick way of doing the four number case without refering to the three case :). $\endgroup$ – Cosmin Pohoata Feb 12 '12 at 5:10
  • $\begingroup$ You're welcome. You saw that by now M.Stoll has completely solved the 4-term problem, even over the rationals. For the 3-term problem, you write that you "checked manually" up to $10^9$; how did you do that? I used the factorization of $t^2-1$ in $bc+1=t^2$ to search up to $t = 10^8$ in a few hours, still finding only the Pell solutions up to $(a,b,c) = (7865521, 58709048, 109552575)$; so $t=10^9$ (if that's what you meant) is feasible too, though it would take a while (or a number of CPU's running in parallel) to finish − and it's certainly not "manual" in the usual sense of "by hand"... $\endgroup$ – Noam D. Elkies Feb 12 '12 at 19:56

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.

  • $\begingroup$ This suggests that indeed the rational points are likewise fully accounted for by finitely many curves. The question asked only for integer points, which could conceivably be more tractable, though I don't see how. $\endgroup$ – Noam D. Elkies Feb 11 '12 at 21:39
  • $\begingroup$ @Noam: The problem with three terms should give a K3 Surface. Are there any K3-related methods that might prove that (up to signs) all the integral points are on the curve that comes from the Pell equation? $\endgroup$ – Michael Stoll Feb 11 '12 at 22:01
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    $\begingroup$ There are no solutions with five terms $a,b,c,d,e$ in arithmetic progression, since then $ab+1$, $ac+1$, $ad+1$, $ae+1$ would have to be four squares in arithmetic progression. Of course, this doesn't really help... $\endgroup$ – Michael Stoll Feb 11 '12 at 22:23
  • $\begingroup$ @Michael: It might be an interesting K3 surface but I don't see how to use this to prove anything about its integral points (for which it's of "log-general type"). $\endgroup$ – Noam D. Elkies Feb 12 '12 at 20:39

Dujella has written many papers on Diophantine m-tuples, check out his webpage.


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.

  • $\begingroup$ To relate this to Michael's answer, Fermat's four square problem is well known to equivalent to computing $X_0(24)(\mathbf{Q})$. To see the equivalence, if $1$, $1+t$, $1+2t$, and $1+3t$ are squares, then one has a rational point on $s^2 = (1+t)(1+2t)(1+3t)$ which is $X_0(24)$. The converse is slightly more subtle; one has to take points in the corresponding coset of the $2$-Selmer group. $\endgroup$ – Shadow of Felipe Jan 15 '16 at 5:12

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 ).

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    $\begingroup$ To complete this: From $d \ge 4bc$, it follows easily that $a,b,c,d$ cannot be an AP. Note also the following: Acdording to Lemma 13 in Dujella's paper linked to in his answer above, if $a,b,c$ is a Diophantine triple with $a < b < c$, then $c = a + b + 2 \sqrt{ab+1}$ or $c \ge 4ab$. If $a,b,c$ form an AP, then the second possibility cannot occur, and the first (together with the AP condition) then implies that $b$ is even and $y^2 - 3(b/2)^2 = 1$, where $y = b-a = c-b$. So it is indeed the case that all such triples come from this Pell equation. (This gives another proof.) $\endgroup$ – Michael Stoll Feb 22 '12 at 19:21

I think it is better to solve a more formal task. We write the system.


We need to find solutions $a,b,c$ - that was an arithmetic progression. This will help the solution of the equation Pell.


Knowing any solution of the equation Pell $(p_0;s_0)$ you can always find the following formula.



Having any decision - can immediately write down the solution of this system.







Interesting here is that the $x,y,z$ looks like an arithmetic progression.


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