Quadratic Diophantine Equations I note that the Diophantine equation, $x^2 + y^2 = z^2$, with $x, y, z \in \mathbb{N}$, has infinitely many solutions. Indeed, $(x, y, z) = (3,4,5)$ provides a solution, and for any $k \in \mathbb{N}$ : $(kx, ky, kz ) = (3k, 4k, 5k)$ provides a solution.
However, assuming $x, y, z \in \mathbb{N}$ with $x, y > 1$, is the same true for the Diophantine equations, 
$x^2 +y^2 = z^2 + 1$, 
$x^2 + y^2 = z^2 + 2$,
$x^2 + y^2 = z^2 + 3$
and more generally, for $x^2 + y^2 = z^2 + n$, for any $n \in \mathbb{N}$?
In particular, are there infinitely-many triples $(x, y, z) \in \mathbb{N}^3$ for which  $x^2 + y^2 = z^2  + n$ is true for infinitely-many values of $n \in \mathbb{N}$?

EDIT: I can't quite parse the final question in the original post. However, given any integer $N,$ there are infinitely many triples $(x,y,z)$ that solve
$$ x^2 + y^2 - z^2 = N^2  $$ using the process below. A "seed" triple may be taken with any $x=z, \; y = N.$ If we just take $\gcd(x,N) = 1$ as well, we get primitive triples. 
ORIGINAL:Actually, the group of automorphs of $x^2 + y^2 - z^2$ is known. It is infinite, and generated by three matrices, along with negating any of $x,y,z$ if that is necessary, not sure. I am trying to find a question that shows the matrices, no luck so far. However, what is means is that, if there is a single solution to $x^2 + y^2 = z^2 + k$ for some integer $k,$ positive or negative or zero, then there are infinitely many, and we can travel among them by matrix multiplication of the column vectors $(x,y,z)^T.$
I think it was some question here or on MO more likely, about the structure of Pythagorean triples. Note that the expert on this is named Ian Agol. There is just a comment by him at this one:
https://mathoverflow.net/questions/33697/assistance-with-understanding-parent-child-relationships-in-pythagorean-triples 
Here we go, http://en.wikipedia.org/wiki/Tree_of_Pythagorean_triples
these matrices due to F. J. M.  Barning (1963)
$$  A = \;
 \left(  \begin{array}{rrr}
  1 & -2 & 2    \\
  2 & -1 & 2 \\
   2 & -2 & 3
\end{array} 
  \right)  
\; \; B = \; 
 \left(  \begin{array}{rrr}
  1 & 2 & 2    \\
  2 & 1 & 2 \\
   2 & 2 & 3
\end{array} 
  \right)  
\; \; C = \; 
 \left(  \begin{array}{rrr}
    -1 & 2 & 2    \\
  -2 & 1 & 2 \\
   -2 & 2 & 3  
\end{array} 
  \right) 
  $$
