Maybe not hard,but I spend 1 hour to solve it..... We know that there exist infinitely many solution for the Pell's Equation: $v^2-3u^2=1$ Let $a=v+2u,b=-v+2u,c=2u$,(we can check that $a,b,c>0$) then $ab+1=4u^2-v^2+1=u^2$, $bc+1=4u^2-2uv+1=u^2-2uv+v^2-1+1=(u-v)^2$, $ac+1=4u^2-2uv+1=u^2+2uv+v^2-1+1=(u+v)^2$

I found an expression which gives infinitive sln. $ (n^2-1,n^2+2n,1)$ $ (n^2-1)(n^2+2n)+1=(n^2+n-1)^2$ $ (n^2-1)(1)+1=n^2$ $ (n^2+2n)(1)+1=(n+1)^2$

But those aren't even in arithmetic progression?? My idea: \[ (a, 4m(4ma-1), 4n(4na+1) )\] We have three variables, in order to tinker: 1. the expressions in an arithmetic progression 2. solve the pell eq. $ 4mn(4ma-1)(4na+1) +1 = k^2$ This is possible but messy unless mods start messing things up. In that case play around with the +/- 1 in the second two of the three numbers ...

me@home // Hawk Tiger's solution is almost right. It just has a little mistake. In his solution, (b,c,a) (not (a,b,c)) is arithmetic progreesion. (b=2u-v, c=2u, a=2u+v)

since the pell's equation gives us infinite number of positive integers $u,v$ such that $v^{2}-3u^{2}=1$, you can easily check that : $a=u$ ,$b=4u+2v$,$c=7u+4v$ works, as we have : $ab+1=(u+v)^{2},$ $bc+1=(5u+3v)^{2},$ $ca+1=(2u+v)^{2}.$

Hawk Tiger wrote: Maybe not hard,but I spend 1 hour to solve it..... We know that there exist infinitely many solution for the Pell's Equation: $v^2-3u^2=1$ Let $a=v+2u,b=-v+2u,c=2u$,(we can check that $a,b,c>0$) then $ab+1=4u^2-v^2+1=u^2$, $bc+1=4u^2-2uv+1=u^2-2uv+v^2-1+1=(u-v)^2$, $ac+1=4u^2-2uv+1=u^2+2uv+v^2-1+1=(u+v)^2$ I understand the latter half of your solution, mainly once we have a, b and c, we can check to see if they are part of an arithmetic progression and that ab+1, bc+1 and ca+1 are perfect squares. However, I am confused about how your solution guarantees an infinite number of triples that meet these requirements. Can you clarify the connection between your Pell Equation and your a, b, and c triple? Thank you

Could someone explain the motivation for the above solution? I understand fully, but where would you get the motivation to use the Pell Equation $v^2-3u^2=1$?

I am also interested in the motivation.

Are these the only triples that satisfy the conditions?

Well a crude Python search up to $10000$ motivated me to consider the Pell Equation $x^2-3y^2=1$.

By Lemma 13 here, if $a<b<c$ and all of $ab+1.bc+1,ca+1$ are perfect squares, then either $c = a + b + 2 \sqrt{ab+1}$ or $c \ge 4ab$. If $a,b,c$ form an arithmetic progression, 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 $y=b-a=c-b$. Hence all such triples come from the Pell equation. This implies that there are no arithmetic progressions of length $4$ satisfying a similar condition.

hamup1 wrote: Could someone explain the motivation for the above solution? I understand fully, but where would you get the motivation to use the Pell Equation $v^2-3u^2=1$? I am not entirely sure if this constitutes as motivation, but the goal is to find some parameterization of (a, b, c) in terms of polynomial expressions. (In our case, that would be polynomials in u and v, which are specifically a=u, b=4u+2v, and c=7u+4v). The existence of an infinite number of solutions for (a,b,c) is then analogous to an infinite number of (u,v) pairs. "Infinite solutions" and "polynomials" triggers Pell's equations, since this imposes favorable constraints that can help manipulate (u, v) to their desired form (that is ab+1, bc+1, ac+1 being squares). At that point you can start searching for linear expressions in u, v for a, b, c and some D for which $u^2-Dv^2=1$. Keep in mind you want things to be as simple as possible. Following these guidelines does eventually get you to a solution. Hope that helps.

hamup1 wrote: Could someone explain the motivation for the above solution? I understand fully, but where would you get the motivation to use the Pell Equation $v^2-3u^2=1$? Another possible motivation: Firstly, to utilize the AP condition, we make the standard substitution $(a,b,c) \mapsto (a-v,a,a+v)$, an then we have that

$a \mapsto 2u$. Then some wishful thinking gives

in the first 2 expressions also appears in the last expression! So, quite naturally, we try and set $3u^2-v^2+1=0$, and this is, fortunately, a form of Pell's equation!

Hawk Tiger wrote: Maybe not hard,but I spend 1 hour to solve it..... We know that there exist infinitely many solution for the Pell's Equation: $v^2-3u^2=1$ Let $a=v+2u,b=-v+2u,c=2u$,(we can check that $a,b,c>0$) then $ab+1=4u^2-v^2+1=u^2$, $bc+1=4u^2-2uv+1=u^2-2uv+v^2-1+1=(u-v)^2$, $ac+1=4u^2-2uv+1=u^2+2uv+v^2-1+1=(u+v)^2$ In your solution I see that $ 2b \neq a+c $

inom wrote: I found an expression which gives infinitive sln. $ (n^2-1,n^2+2n,1)$ $ (n^2-1)(n^2+2n)+1=(n^2+n-1)^2$ $ (n^2-1)(1)+1=n^2$ $ (n^2+2n)(1)+1=(n+1)^2$ In this case $ 2b \neq a+c $

Tuhin_Bose wrote: In your solution I see that $ 2b \neq a+c $ But $2c=a+b$