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

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b=a+1, c=ab+1 gives an infinite class of solutions. This can be found from (a+i)(a-i)(b+i)(b-i)=(c+i)(c-i) by choosing (a-i)(b+i)=c+i. I do not know whether there are other solutions.^2%2B1%29*%28a^2%2B2*a%2B2%29-%28a^2%2B‌​a%2B1%29^2 – Martin Sleziak Apr 9 '11 at 11:50
This problem is really bugging me, if $(c+i)(c-i) = x \bar{x}$ we can't say that $x = c\pm i$ so it seems like Gauss integers are basically useless for this problem. – quanta Apr 9 '11 at 15:04
Dear Amir, please consider adding some explanation or motivation to your questions. – Akhil Mathew Apr 9 '11 at 17:45
This is also equivalent to $a^2b^2+a^2+b^2=c^2$. I.e., (ab,a,b,c) is a Pythagorean quadruple. Several parametrizations of Pythagorean quadruples (x,y,z,w) are known, but I do not think that adding the condition x=yz will bring something reasonable. – Martin Sleziak Apr 10 '11 at 8:35
@Martin, a couple of smallish solutions which aren't $b=a+1$ are $1,12,17$ and $2,8,18$. – Gerry Myerson Jul 18 '11 at 3:54

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.

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Links to the paper: doi: 10.1007/BF03026559, GDZ. (I did not want to bump the post so I've added them in the comment instead.) – Martin Sleziak Jun 25 '12 at 13:50

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