Finding solutions if existed in this number theory - prime numbers problem. I was working on a problem involving prime number and I reached a point where I could not solve an equation, and the equation is.
Given that $a,b$ and $c$ are positive integers which satisfies that both $a$ and $b $ are less than $c$.
And satisfy the following equation
$$ac=b^2 +1$$
Find solutions if theres any, if there exists infintly many solutions find $a$ and $ab$ as functions in tearms of $c$.
(The original problem was if given a number $s$, find the biggest prime divisor for ($s^2 +1$). And after some conclusions I came up with the above equation and didnt know how to find a solution, so if the first problem cannot be answered I hope that the second one can be).
 A: Changing your letters to become $$ y^2 - z x = -1. $$
The automorphisms (with positive determinant) of the quadratic form $y^2 - zx$ are known, and parametrized by the modular group. So, given $ps-qr = 1,$ so
$$
\det
\left(
\begin{array}{cc}
p & q \\
r & s \\
\end{array}
\right) = 1
$$ 
we construct
$$
W =
\left(
\begin{array}{ccc}
p^2 & 2pq & q^2 \\
pr & ps+qr& qs \\
r^2 & 2rs & s^2 \\
\end{array}
\right)
$$
The Hessian matrix of the quadratic form is
$$
H =
\left(
\begin{array}{ccc}
0 & 0 & -1 \\
0 & 2 & 0 \\
-1 & 0 & 0 \\
\end{array}
\right)
$$
and we have arranged $$  W^T H W = H.  $$
This means that, if
$$
\left(
\begin{array}{c}
x \\
y  \\
z \\
\end{array}
\right)
$$
solves $y^2 - zx = -1,$ so does every
$$
\left(
\begin{array}{ccc}
p^2 x + 2pq y + q^2 z \\
pr x + (ps+qr) y + qs z \\
r^2 x +2rs y + s^2 z \\
\end{array}
\right)
$$
For example, if we take $x=1, y=0, z=1$ to begin, 
$$
\left(
\begin{array}{ccc}
p^2  + q^2  \\
pr  + qs  \\
r^2  + s^2  \\
\end{array}
\right)
$$
is a solution vector whenever $ps-qr = 1$
