On the equation $(a^2+1)(b^2+1)=c^2+1$

How do you find all positive integers $a,b,$ and $c$ such that $(a^2+1)(b^2+1)=c^2+1$?

Solution 1:

See Kenji Kashihara, Explicit complete solution in integers of a class of equations $(ax^2−b)(ay^2−b)=z^2−c$, Manuscripta Math. 80 (1993), no. 4, 373–392, MR1243153 (94j:11031).

The review in Math Reviews says,

"The author studies the Diophantine equation $(ax^2−b)(ay^2−b)=z^2−c$, where $a,b,c\in{\bf Z}$, $a\ne0$, $b$ divides 4, and in the case $b=\pm4$, then $c\equiv0\pmod 4$. This equation for $a=1, b=1$ has been treated by S. Katayama and the author [J. Math. Tokushima Univ. 24 (1990), 1--11; MR1165013 (93c:11013)], and the present paper extends the techniques to show that there exists a permutation group $G$ on all integral solutions of the equation, and also an algorithmic method for computing a minimal finite set of integral solutions, in the sense that all integral solutions are contained in the $G$-orbits of the set. Such minimal sets are listed for the equations with $a=2, b=\pm1, 0\lt|c|\le85$."

Looks like this includes the case $a=1$, $b=-1$, $c=-1$ which is what we want.