Integral solutions of $x^2+y^2+1=z^2$ I am interested in integral solutions of $$x^2+y^2+1=z^2.$$ Is there a complete theory comparable to the one for $x^2+y^2=z^2?$
The equation defines a conic with a rational point (0,0,1). The other rational solutions can be parametrized by lines through that point.  As with $x^2+y^2 = z^2$ the integer solutions can be deduced from the rational ones, and the quadratic form $x^2+y^2 - z^2$ has a large linear symmetry group allowing one to move between solutions.  In these respects the theory is the same as the one for Pythagorean triples.

There is a difference in the structure of the symmetry group.  For $a=0$ the hyperboloid $x^2+y^2 - z^2 = a$ becomes a cone, with additional scaling symmetries in addition to the linear transformations of the hyperboloid; in effect the $O(2,1)$ symmetry group of the rational solutions collapses into a product of scalings and circle isometries.  For integer solutions, with Pythagorean triples there is a reduction to the case of primitive triples, but when $a \neq 0$ there is a bound on the shared factors of $(x,y,z)$, and for $a = \pm 1$ all integer solutions are primitive.  The organization of Pythagorean triples using several 3x3 integer matrices as transformations connecting different solutions does use the $O(2,1)$ structure, and the solution set for the equation with $a \neq 0$ could be presented in the same way (possibly with more than one connected component when $|a| > 1$).
All of numbers can be any character.In Equation:  $qX^2+Y^2=Z^2+a$  

If the ratio is factored so:  $a=(b-c)(b+c)$ 

Then we use the solutions of Pell's equation: $p^2-fs^2=\pm1$ 

where:  $f=(q+1)k^2-2kt-(q-1)t^2$ 

Then the solutions are of the form: 

$X=2(ck-bt)ps+2(bk^2-(b+c)kt+ct^2)s^2$ 

$Y=bp^2+2c(k-t)ps-(b(q-1)k^2+2(b-qc)kt+b(q-1)t^2)s^2$  

$Z=cp^2+2b(k-t)ps+(c(q+1)k^2-2(bq+c)kt+c(q+1)t^2)s^2$ 

All of numbers can be any character.