Prove that there are infinite sets of $(a, b, c)$ such that $ab + 1$, $ac + 1$ and $bc + 1$ are perfect square Prove that there are infinite sets of integer numbers $(a, b, c)$ that $ab + 1$, $ac + 1$ and $bc + 1$ are perfect square. ($a$, $b$, $c$ are different numbers)
Any hints how to prove the statement? I'm trying to find a way to turn that into a Pell's equation
and then prove it has infinite answers.

If $a=4n$ , $b=n+1$ and $c=n-1$ then
\begin{eqnarray*}
ab+1&=&(2n+1)^2 \\
bc+1&=& n^2 \\
ca+1&=&(2n-1)^2
\end{eqnarray*}
EDIT : for Ovi.
We $ab+1,bc+1,ca+1$ are perfect squares. So there exist $x,y,z$ such that
\begin{eqnarray*}
z^2=ab+1 \\
x^2=bc+1 \\
y^2=ca+1
\end{eqnarray*}
Now subtract these equations pairwise (difference of two squares factorise) and we have
\begin{eqnarray*}
(z-x)(z+x)=b(a-c) \\
(z-y)(z+y)=a(b-c) \\
(y-x)(y+x)=c(a-b)
\end{eqnarray*}
At this point I totally guessed to equate the factors pairwise (there is no justification for this). Amazingly this gives 
\begin{eqnarray*}
x=\frac{-a+b+c}{2} \\
y=\frac{a-b+c}{2} \\
z=\frac{a+b-c}{2} 
\end{eqnarray*}
Now substitute this formula for $z$ into $z^2=ab+1$ & after some algebra, we have
\begin{eqnarray*}
a^2+b^2+c^2-2(ab+bc+ca)=4.\\
a^2-2a(b+c)+(b-c)^2-4=0
\end{eqnarray*}
Now consider this as a quadratic in $a$ ... Its disciminant (over 4) needs to be a perfect square ...
\begin{eqnarray*}
\Delta=(b+c)^2- \left( (b-c)^2-4 \right) =4bc+4=(2n)^2.
\end{eqnarray*}
& the above solution follows.
