Clever use of Pell's equation Find infinitely many triples $(a,b,c)$ of positive integers such that $a,b,c$ are in arithmetic progression and such that $ab+1$, $bc+1$, and $ca+1$are perfect squares.
The solution is:
Consider the Pell's equation $x^{2}-3y^{2}=1$. If $(r,s)$ is a solution, then the triple 
$(a,b,c)=(2s-r,2s,2s+r)$ is in arithmetic progression and $(2s-r)(2s)+1 = (r-s)^{2}$,
$(2s-r)(2s+r)+1 = s^{2}$, and $(2s)(2s+r)+1 = (r+s)^{2}$
Can someone explain to me how the author came up with this solution? How did he know to use $x^{2}-3y^{2}=1$?

(Old answer revised.)

I. Method. 

Here is a sketch of how one can find that Pell equation (and others) from first principles. Let,
$$\begin{aligned}
ab+1\;& = (p_1x+p_2y)^2\\
ac+1\;&  = (p_3x+p_4y)^2\\
bc+1\;&  = (p_5x+p_6y)^2
\end{aligned}\tag1$$
Since we wish $a,b,c$ to be in arithmetic progression, assume,
$$a,\,b,\,c = -q_1x+q_2y,\;q_3y,\;q_1x+q_2y$$
for unknown integers $p_i,\,q_i$.Expand $(1)$ and collect powers of $x,y$,
$$\begin{aligned}
r_1x^2+r_4xy+r_7y^2 \;&  =1\\
r_2x^2+r_5xy+r_8y^2 \;&  =1\\
r_3x^2+r_6xy+r_9y^2 \;&  =1\\
\end{aligned}$$
where the $r_i$ are in terms of the $p_i,\,q_i$. The above is a clue that a Pell equation may be involved. Then solve the system,
$$\begin{aligned}
& r_1 = r_2 = r_3=1\\
& r_4 = r_5 = r_6=0\\
& r_7 = r_8 = r_9\\
\end{aligned}$$
which is quite easy to do. 

II. Solution. 

We find,
$$\begin{aligned}
ab+1\;& = \big(x-(m+n)y\big)^2\\
ac+1\;& = (ny)^2\\
bc+1\;& = \big(x+(m+n)y\big)^2
\end{aligned}$$
where $a,\,b,\,c = -x+my,\;2(m+n)y,\;x+my,\,$ and $x,y$ solve the more general Pell equation,
$$x^2-(m^2-n^2)y^2=1\tag2$$
Since $a,b,c$ are to be in arithmetic progression, let $m = 2(m+n)$, so $m,n = 2,-1$ and  $x^2-3y^2=1$ pops out. 

III. Others. 

Using the same method,
$$\begin{aligned}
ab-1\;& = (my)^2\\
ac-1\;& = \big(x+(m+n)y\big)^2\\
bc-1\;& = \big(x+(m-n)y\big)^2
\end{aligned}$$
where $a,\,b,\,c = x+ny,\;x-ny,\;2(x+my),\,$ and $x,y$ solve the similar Pell equation,
$$x^2-(m^2+n^2)y^2=1\tag3$$

IV. Higher Powers

Also by solving a system of equations, given,
$$x^2-17y^2 =\pm1\tag2$$
then,
$$(13x^2 + 12x y - 17y^2)^4 + (13x^2 - 12x y - 17y^2)^4 = (239x^4 - 14x^2 y^2 - 289y^4)^2+1$$
As you may notice, Pell equations are useful with other Diophantine equations where one term is set equal to 1.
