Integral solutions of hyperboloid $x^2+y^2-z^2=1$ Are there integral solutions to the equation $x^2+y^2-z^2=1$?

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.  
