A question essentially the same a this one was asked in MSE 4479792 without any background details and was deleted by the post author after I posted a minimal one sentence answer mentioning five OEIS sequences (more like a long comment).
I propose the question again but including more attention to details and background.
Prove that all of the positive integer solutions of
$$ ab\!+\!1\!=\!x^2,\: ac\!+\!1\!=\!y^2,\: bc\!+\!1\!=\!z^2,\: x+z\!=\!2y,\; 0\!<\!a\!<\!b\!<\!c $$
are given by
$$ a=A001542(n),\quad b=A005319(n),\quad c=A001542(n\!+\!1),\\ x=A001541(n),\quad y=A001653(n\!+\!1),\quad z=A002315(n) $$
with $0<n.$ These six sequences all satisfy the same linear recurrence relation but with different initial values. For example, $\,a_n = 6\,a_{n-1}-a_{n-2}\,$ and similarly for the other five sequences. Notice the condition that $\,x+z=2y\,$ is equivalent to the condition that $\,x,y,z\,$ are in arithmetic progression. Without this condition the problem is a special case of a problem in Diophantus according to L. E. Dickson, History of the Theory of Numbers, Volume II, Chapter XIX, Section "$xy+a, xz+a, yz+a$ all squares", pp. $513$-$515$
Diophantus, III, $12, 13$ and IV, $20$, asked for three numbers such that the product of any two increased by a given number $a$ shall be a square. For $\,a=12,\,$ he found $\,2,2,1/8;\,$ for $\,a=-10,\,$ complicated fractions; for $\,a=1, x, x+2, 4x+4.\,$
On page $515$ Dickson states
Fr. Buchner$^{83}$ treated $xy+1=p^2, xz+1=q^2, yz+1=r^2.$
however both Diophantus and Buchner allow rational solutions and there is no explicit mention of arithmetic progression in Dickson for this Diophantine system.
I wonder what is the original source of the arithmetic progression condition of the square roots for this problem?
An answer to this question requires:
- A1. A proof that the sequences give a solution.
- A2. A proof that there are no other solutions.
A proof of A1 is elementary given the recursions and induction, however a proof of A2 is much harder, but the solutions all depend on the Pell equation $\,x^2-2a^2=1\,$ and so results from that theory probably need to be used in the proof.
For convenience and as a check, a small table of sequence values:
$$ \begin{array}{|r|r|r|r|r|r|r|} \hline n && a_n & b_n & c_n && x_n & y_n & z_n \\ \hline -3 && -70 & -140 & -12 && 99 & 29 & -41 \\ \hline -2 && -12 & -24 & -2 && 17 & 5 & -7 \\ \hline -1 && -2 & -4 & 0 && 3 & 1 & -1 \\ \hline 0 && 0 & 0 & 2 && 1 & 1 & 1 \\ \hline 1 && 2 & 4 & 12 && 3 & 5 & 7 \\ \hline 2 && 12 & 24 & 70 && 17 & 29 & 41 \\ \hline 3 && 70 & 140 & 408 && 99 & 169 & 239 \\ \hline \end{array}$$
NOTE: all four algebraic conditions are satisfied for all integer $\,n\,$ but $\,0\!<\!a_n\!<\!b_n\,$ requires $\,0\!<\!n.$
ADDED: For convenience and to save some wasted effort, I give an elementary proof of A1 here using only some simple algebra of equations and inequalities. First,
$$ b=2a,\quad c=3a+2x,\quad y=a+x,\quad z=2a+x$$
are proved true since they are satisfied by the initial values for $\,n=1,2\,$ and are preserved by the common linear recursion relations. The last two equations combined prove the important $\,x+z=2y\,$ arithmetic progression condition. Also,
$$ a_{n+1} = 3a_n+2x_n,\quad b_{n+1} = 6a_n+4x_n,\quad x_{n+1} = 4a_n+3x_n $$
are proved the same way. Given the initial values $\,0<a_1=2<x_1=3,\,$ and given the previous results, this proves $\, 0<a_n<x_n<b_n<y_n<z_n<c_n\, $ for all $\,n>0.\,$ Use previous results to get
$$ ab+1-x^2 = ac+1-y^2 = bc+1-z^2 = 2a^2 + 1 - x^2. $$
But $\,2a_0^2+1-x_0^2 = 0\,$ and previous results give $\,2a_{n+1}^2+1-x_{n+1}^2 = 2a_n^2+1-x_n^2\,$ which implies this is $0$ for all $\,n.$ This proves
$$ ab\!+\!1=x^2,\quad ac\!+\!1=y^2,\quad bc\!+\!1=z^2. $$