How to find general solution (in terms of parameters) for diophantine equations $x^2+y^2-z^2=1$ and $x^2+y^2-z^2=-1$?
It's easy to find such solutions for $x^2+y^2-z^2=0$ or $x^2+y^2-z^2-w^2=0$ or $x^2+y^2+z^2-w^2=0$, but for these ones I cannot find anything relevant.
I think that the solutions to $x^2+y^2-z^2=-1$ are $x=RT-SU,y=RU+ST$ where $R^2+S^2-T^2-U^2=2$ then $z=R^2+S^2-1=T^2+U^2+1$ On the surface this looks similar to the solutions to the $+1$ case. However these are quite a bit rarer and depend on the locations of the primes.
As we know, an integer can be uniquely written as $n=ab^2$ where $a$ (the squarefree part of $n$) is a product of distinct primes. $n$ can be written as a sum of two squares $n=j^2+k^2$ precisely when $a$ has no prime divisors of the form $4m+3$ (and we know in how many ways this can be done as well.) So the solutions depend on when we have 2 consecutive even numbers of this form.
For example $292=73\cdot4^2$ and $290=2\cdot5\cdot329$ thus we know that there are expressions as a sum of two squares: $$292=6^2+16^2$$ $$290=1^2+17^2=11^2+13^2.$$ Running through the various possiblities gives these solutions for $R,S,T,U,x,y$ with $x^2+y^2-291^2=-1:$
Certain families of solutions can be given. One is $x,y,z=2p,2p^2,2p^2+1.$
The question is the same as looking for points of norm 1 or -1 in the unimodular Lorentzian lattice $Z^{1,2}$. This has an infinite group of automorphisms, with an index 2 subgroup that is a Coxeter group generated by 3 reflections. This group acts transitively on the vectors of norm 1 and -1 if I remember correctly, so all solutions can be obtained from 1 particular solution by acting with this group.
I believe the general solution to $x^2+y^2-z^2=1$ is $x=(rs+tu)/2$, $y=(rt-su)/2$, $z=(rs-tu)/2$, where $rt+su=2$.
EDIT: Solutions to $x^2+y^2+1=z^2$ can be obtained by choosing $a$, $b$, $c$, $d$ such that $ad-bc=1$ and then letting $x=(a^2+b^2-c^2-d^2)/2$, $y=ac+bd$, $z=(a^2+b^2+c^2+d^2)/2$, though I'm not sure you get all the integer solutions this way.
The rational solutions are a bit easier. $(0,0,1)$ is a (rational) point on the surface. The line $(0,0,1)+t(a,b,c)$ through that point intersects the surface again at $x=2ac/(a^2+b^2-c^2)$, $y=2bc/(a^2+b^2-c^2)$, $z=(a^2+b^2+c^2)/(a^2+b^2-c^2)$, giving all the rational points on the surface.
Let´s take any $x>3$ and choose a,b such that $a<b$,$a-b$ even and $ab=x^2-1$. Then with $y=(b-a)/2$, $z=(b+a)/2$ we have $x^2 + y^2 = z^2 + 1$. Particular cases∶ if $x$ even then $x^2 + ((x^2-2)/2)^2 = (x^2/2)^2 + 1$; if $x$ odd then $x^2 + ((x^2-5)/4)^2 = ((x^2+3)/4)^2 + 1$.
Of course for the equation $X^2+Y^2=Z^2+t$
There is a particular solution:
$X=1\pm{b}$
$Y=\frac{(b^2-t\pm{2b})}{2}$
$Z=\frac{(b^2+2-t\pm{2b})}{2}$
But interessuet is another solution:
$X^2+Y^2=Z^2+1$
If you use the solution of Pell's equation: $p^2-2s^2=\pm1$ Making formula has the form:
$X=2s(p+s)L+p^2+2ps+2s^2=aL+c$
$Y=(p^2+2ps)L+p^2+2ps+2s^2=bL+c$
$Z=(p^2+2ps+2s^2)L+p^2+4ps+2s^2=cL+q$
number $L$ and any given us. The most interesting thing here is that the numbers $a,b,c$ it Pythagorean triple.
$a^2+b^2=c^2$
This formula is remarkable in that it allows using the equation $p^2-2s^2=\pm{k}$ Allows you to find Pythagorean triples with a given difference.
$a=2s(p+s)$
$b=p(p+2s)$
$c=p^2+2ps+2s^2$
$b-a=\pm{k}$
Pretty is not expected relationship between the solutions of Pell's equation and Pythagorean triples.
